Common Functions - Online Tutor, Practice Problems & Exam Prep

Graphs of Common Functions

Hey, everyone. So up to this point, we've been spending a lot of time focusing on functions. We've been looking at the graphs of functions and finding the domain and range. And in this video, we're going to be taking a look at some of the common functions. These are the functions that are going to frequently show up throughout this course.

It's definitely important to make sure we're familiar with these functions and their graphs and just in general how they work. Let's take a look at the constant function. The constant function occurs when our function \( f(x) = c \), where \( c \) can be any constant number. For example, our function \( f(x) = 2 \) would be a constant function because 2 is just a constant, and notice on our graph that \( y = 2 \) is constantly the value in all directions. We can input any \( x \) that we want into this function. So, we would say that the domain goes from negative infinity to positive infinity because we can have any \( x \) value we want. But notice how the range is only going to be where our \( y \) value is 2. \( y = 2 \) is the only possible output we can get for this function. Remember, this value is just going to be whatever \( c \) is, so it's not always going to be 2; it could be 4 or negative \( \frac{5}{3} \).

Now let's take a look at the identity function next. The identity function says that \( f(x) = x \), and this function tells you that whatever you put into the function, you're going to get out of it. For example, if you put negative one in for \( x \), you're going to get negative one as your output. Similarly, if you put 50 in for \( x \), you're going to get 50 as your output. Because of this, we can say that the domain is all real numbers because we can put any number we want into it. Additionally, the range is also all real numbers.

Now let's take a look at the square function. The square function forms this interesting shape called a parabola. The parabola is this bowl-like shape that you see on the screen here. The square function happens when \( f(x) = x^2 \), and the domain for this function is going to be all real numbers. Note that even though all the positive \( y \)'s are included, none of these negative \( y \)'s are encompassed; the curve does not extend down below the graph. So, we would say that the range will go from 0 all the way to positive infinity, and we do include the 0 because we could have a value right at the origin, but we can never be below 0, so we can never get a negative output for the square function.

Next, let's take a look at the cube function. The cube function is defined by \( f(x) = x^3 \). For this graph, we have all the negative \( x \)'s and all the positive \( x \)'s included. So we would say that the domain is all real numbers. Similarly, for the \( y \) values, we include all the negative and positive \( y \)'s in this curve. So, the range is also all real numbers.

Now let's take a look at the square root function. And the square root function has the most restrictions of all the common functions. In this function, we look at our graph; notice how this graph continuously goes to the right and also goes up. Because it continuously goes to the right, we have all positive \( x \)'s, but none of the negative \( x \)'s are included. So, we would say that the domain goes from 0, including 0, all the way up to positive infinity. Looking at our range, notice that our range continuously goes up, so we have all the positive \( y \) values, but none of the negative \( y \) values are included. So the range also goes from 0 to positive infinity.

Next, let's take a look at the cube root function. The cube root function is where we have our function \( f(x) = \sqrt[3]{x} \). Notice for this function, we have all of our negative \( x \)'s and all of our positive \( x \)'s included. Thus, the domain goes from negative infinity to positive infinity, as all real numbers are going to be included for the domain because we see that our curve extends to all these values. Similarly, considering our \( y \) values, we can see that all the positive \( y \)'s and all the negative \( y \)'s will also be included because this graph continues to extend downwards and upwards as well as to the left and right. So, the range is going to be all real numbers also. This function does include negative \( x \) values because the negative \( x \)'s are represented in this graph.

So those are some of the common functions that you'll see throughout this course and also in future math courses. Hopefully, you found this INTRODUCTION helpful, and let me know if you have any questions.

Slope-Intercept Form

Hey, everyone. So throughout our discussion on lines, we've seen problems where they ask us to calculate the slope of a line by looking at the rise over the run between two points. But some problems like the one we're going to work out down below won't ask for just that. Some problems will ask us to look at the graph of a line and write its equation in a very specific way called the slope-intercept form. So that's what I want to talk about in this video.

And what I'm going to show you is it's basically just a very specific way that we write an equation of 2 variables using 2 things that we've already seen before independently. So we can use its slope and its intercept, specifically the y-intercept. Alright? So I'm going to show you this equation, and it's probably something that you've seen before. So let's get started.

So the slope-intercept form of an equation is actually just an equation. That's y=mx+b. You've probably heard that at some point in a math class, y=mx+b. This is just the slope-intercept form of a line. It's one of the most easy and straightforward ways to describe a line equation.

So let's get started and talk about these two variables here. The m is basically just the slope, and we've already seen that before. We calculate that by using delta y over delta x, rise over the run, or you can just use the sort of longer format here. So in this equation, what we can see is that the rise over the run between these two points is 2 over 1, and so the slope is just 2. What about the y-intercept?

Well, the b term over here is the y-intercept, and we've already talked about that separately. The y-intercept is basically just the y-value wherever the graph crosses the y-axis, and it's where x is equal to 0. So, for instance, in this case, the graph crosses the y-axis right over here. The y-value at this point is 3. So that is the b term.

Right? So that's y intercept. So to put these things together, y=mx+b, all you have to do is the equation here has a slope of 2, so that goes in front of the x, and then the intercept is 3. So the equation of this line is just y=2x+3. That's how to describe this equation in slope-intercept form.

That's really all there is to it. Alright? So that's all there is to slope-intercept form. Let's go ahead and take a look at another example. Alright?

So in this graph below, we're going to identify the slope and the y-intercept, and then we're going to write the equation in slope-intercept form. So remember, slope-intercept form here is just going to be y=mx+b. In order to figure out m and b, we're going to have to take a look at the graph. Alright? So the always the easiest sort of value start with is going to be the b term because you really just have to look at the graph and figure out where does it cross the y-axis.

You don't have to calculate anything. So let's take a look at this graph. Where does it cross the y-axis? Well, it crosses right over here. So what's the y-value where this graph crosses the y-axis?

It's just negative 3. So that's the b. It's just negative 3. That's all there is to it. You don't have to write the ordered pair.

It's just the y-value. Alright? So how do we calculate m? Well, for m, we're going to have to use rise over run. So we're going to m we're just going to calculate this by using delta y over delta x.

Or if we're given 2 points, you could just plug those points in. But, basically, if I'm going to have this point over here, what's the rise of the run to get to the next point? Well, if I take a look at this graph here, what happens is I've got all these different points that are all at the intersections. These little lines here, all these are valid points that I can pick. So what's the rise of the run?

Well, to get from this point to the next one, all I have to do is I have to go up 1 and then over 1. So up 1 over 1. So that's just a rise runoff of just 1. So the slope is 1. I just go 1 over and sorry.

1 up and 1 over. So because the slope is just equal to 1, I basically just plug both of these things into this equation, and my equation is just equal to y=1x+b, and this is going to be negative 3. Just to simplify this, a lot of times, what you'll see here is if there's just a one in front, you don't even write it. So it's just y=x-3. So this is the equation that describes this line in slope-intercept form, y=x-3.

Alright? So, anyway, folks, thanks for watching. Let me know if you have any questions.

In the graph shown, identify the y–intercept & slope. Write the equation of this line in Slope-Intercept form.



• A. $y=\frac23x+1$
• B. $y=-\frac23x+1$
• C. $y=-2x+1$
• D. $y=x+2$

Properties of Parabolas

Hey everyone, we've worked with some basic graphs of functions, and here we're going to focus on one specific type of function, the quadratic function. Now it might seem like a lot of information at first, and this graph behind me might look a little bit intimidating, but don't worry, I'm going to walk you through absolutely everything that you need to know about the graphs of quadratic functions here, and you'll be an expert in no time. So let's go ahead and jump right in. A quadratic function is a polynomial of degree 2 that has a standard form of f(x)=ax2+bx+c. Now, all of these functions here are examples of quadratic functions and you can see that a,b,c can be any real number whether it be a fraction, a negative number, or even 0, so long as a≠0 and our largest exponent is still 2, making it a quadratic function.

Now here we're going to focus on the graphs of quadratic functions so let's go ahead and take a look here. We've worked with the square function before f(x)=x2 and you may remember that this square function was a parabola. Now this is actually going to be the same for all quadratic functions. They are all going to have this curved parabolic shape. Whether it is right side up, upside down, or located anywhere on our coordinate plane, our quadratic functions will always be that same shape of a parabola.

So we're going to look at the different elements of a parabola here, and some are going to be ones that we're familiar with, like the x intercept or the y intercept, but we're also going to work with some that are specific to parabolas, like the vertex or the axis of symmetry. So let's go ahead and get started with our vertex. The vertex of a parabola is either going to be the lowest point or the highest point depending on whether our parabola is opening upward or opening downward. For our square function here, we see that our vertex is right at that origin point and we're always going to write our vertex as an ordered pair, so my vertex is simply (0, 0). Now, for my other function here, it is not at the origin, it is actually at the point (-2, 1), so my vertex is (-2, 1).

Now the other thing that I want to consider with my vertex is whether I'm dealing with a minimum or a maximum point. So again, looking at my square function here, this is the lowest point on my graph of that quadratic function. So it is a minimum point. It's all the way at the bottom, and nothing goes below it on that parabola. Now for my other function, my vertex is all the way at the top, which tells me that I am not dealing with a minimum. I'm actually dealing with a maximum point here. So here, my vertex is (-2, 1), and it is a maximum. Let's go ahead and look at something that we're more familiar with, our x intercept. The x intercept of a graph is anywhere that our graph crosses our x axis, not our y axis, our x axis.

So we're going to look for the points that cross that x axis. And between my two functions here, I have 3 different points that cross my x axis that represent my x intercepts. So let's take a look at our square function. Now there's only one point that crosses the x axis and it is at my origin, so my x intercept is simply (0, 0). Now looking at my other function, there are 2 points that cross the x axis, so I need to account for both of them, (-3, 0) and (-1, 0), as my x intercepts.

Now you'll only ever have 1 or 2 x intercepts, never less or more than that. So let's move on to our y intercept. The y intercept is where a graph crosses the y axis, this time the y axis. So let's look at our square function first. We have this point. This represents our y intercept. Now our y intercept here is again at the origin, simply at (0, 0). And for my other graph, my other function, my y intercept is down here at (-3, 0), so it's simply (-3, 0). Now we've looked at our intercepts and our vertex. Now we want to look at our axis of symmetry.

Now our axis of symmetry is something that is going to be specific to parabolas, and it represents the line that divides our parabola perfectly in half. It is symmetric about that line. So our axis of symmetry is going to perfectly cut it in half. So looking at my square function, it's going to go straight through the middle, and it's actually always going to go straight through our vertex. So here my axis of symmetry is simply the line x equals 0, a vertical line through my vertex.

Now looking at our other function here, if I draw a line that divides this perfectly in half, you're often going to see this actually written with a dotted line, but I've just highlighted them here so that it's easier to see. But here again, we're going straight through that vertex point at x equals -2. So it's always going to be a straight vertical line through our vertex and that represents our axis of symmetry. If I were to fold my parabola in half at that line, it's going to perfectly match up because it is symmetric about that line. Now something else to consider whenever we're looking at graphs of functions is always the domain and the range.

So first, looking at our domain, the domain of all quadratic functions is actually always going to be the same, and it's always going to be negative infinity to infinity, which you might also recognize as being all real numbers. So for both my square function and for my other function here, they are both all real numbers negative infinity to infinity. Now our range is going to depend on whether we have a minimum or a maximum point. So looking first at our square function, we know that we're dealing with a minimum point here, and that looking at that, my y values can go anywhere from that minimum and all the way up beyond that. So whenever I'm dealing with a minimum, my range is always going to go from y minimum, whatever that minimum point is, all the way up to infinity.

So in this case, since I start down at 0, my range is simply 0 to infinity. Now when I'm dealing with a maximum point like I have here on my other graph, it's always going to go from my maximum point down to negative infinity. So in interval notation, I would write that as negative infinity up to my maximum point y maximum. So in this case, it goes from negative infinity up to my y point at y equals 1, and that's my domain and range. Now we want to look at one last thing on our graphs here.

Something that you're going to often be asked when dealing with parabolas is to tell them the interval for which your parabola is increasing and decreasing. Now this is just a fancy way of saying for what values of x is our graph going up and for what values of x is it going down. So let's just take a look at our second function here that's facing downward and decide where it's increasing and decreasing. Now looking at the first side of this, going down here all the way up to my vertex point, we see that our graph is going up. It is increasing.

So the interval for which it is increasing is simply from negative infinity up to that vertex point at x equals -2. And our vertex is actually always going to divide the intervals for which we're increasing or decreasing. So the other side, from x equals -2 all the way to infinity, we're going to be decreasing because we see our graph is going down. So this is just simply the opposite, from x equals -2 to infinity. So that's all the stuff we need to know about parabolas.

One more thing that I want to mention here is that you might have noticed that my equation here doesn't quite look like it's in standard form. And that's actually because we're often going to see quadratic functions written in what's called vertex form, which is what that is right there. And it's actually going to help us to graph with ease. It's going to help us to easily graph these quadratic functions, which is what is coming up next. So thanks for watching.

I'll see you in the next video.

Identify the ordered pair of the vertex of the parabola. State whether it is a minimum or maximum.

• A. $\left(-3,-2\right);$ maximum
• B. $\left(-3,-2\right);$ minimum
• C. $\left(-1,2\right);$ maximum
• D. $\left(-1,2\right);$ minimum

Where is the axis of symmetry located on the given parabola? 

• A. $x=3$
• B. $x=0$
• C. $x=1$
• D. $x=4$

Introduction to Polynomial Functions

Hey, everyone. We've worked with polynomial expressions, and we've even worked with a specific type of polynomial function, a quadratic function, where my highest power is 2. Now I want to take a broader look at all the possibilities of polynomial functions, which can really just be any polynomial, but now f of x is equal to that polynomial, making it a polynomial function. So that might sound like a lot, any polynomial, and now it can be a function too. But don't worry.

We're going to rely on a lot of what we already know about polynomials and some of what we just learned about quadratic functions in order to learn more about polynomial functions and their graphs. So let's go ahead and jump right in. Looking at the polynomial function that I have right here, we want to remind ourselves of a couple of things that we learned with polynomials. The first of which is that polynomials can only have positive whole number exponents. This means no negatives and no fractions in those exponents.

The other thing is that whenever we write our polynomials in standard form, all of our like terms need to be combined, and it needs to be written in descending order of power. So if I start with a power of 3, my next power is going to be 1 lower and then 1 lower, 1 lower until I get to the last one descending order. Now looking at these polynomials that I have here, you might notice this one looks similar to what you've seen in your textbook. And this can look a little bit intimidating when you first see it, but don't worry. This is just showing us exactly how to write any polynomial in standard form.

So this a_n right here represents my leading coefficient. So in the polynomial that I have here, it's simply 6, and n represents the degree of the polynomial. So in my polynomial here, it's simply 3. Now when I look at my next term, I see that this power goes down to n-1, and that's just showing us that it's in descending order of power. So here I started with 3.

If I subtract 1 from that, I get 2, which is my very next power here. And then all of these a terms down to my a₀ just represent all of my coefficients and then my constant. So don't let that form freak you out; it's just showing you how to write in standard form. Okay, so let's just look at some polynomial functions here. Looking at this first one, I have f(x)=-x2+5x3-6x+4 and I want to determine if this even is a polynomial function and if it is, write it in standard form and state both the degree and leading coefficient.

So let's first determine if this even is a polynomial function. And since I only have positive whole number exponents, I don't have any fractions or any negatives in those exponents, it looks like, yes, this is a polynomial function. So let's go ahead and write it in standard form. Now here, it looks like I just need to switch these two terms in order for it to be in standard form. So I end up with f(x)=5x3-x2-6x+4, and we're good.

We're in standard form. So let's go ahead and identify that degree. So the degree of this polynomial, looking at that first term, my highest power, is simply 3, and then my leading coefficient is the one that's attached to that x cubed. It is 5. So we're good on that one.

Let's go ahead and move to our next one. Here we have f(x)=2x12+3. Now is this a polynomial function? Well, I have a fraction in my exponent there, so no, this is not a polynomial function. I cannot have fractions as exponents, so I don't have to worry about anything else because it's not a polynomial function at all.

Let's go ahead and move on to our last example here. We have f(x)=-23x4+1+3. Now looking at this, there is a fraction as a coefficient. Does that mean that it's not a polynomial? Well, no, because it's just a coefficient.

It's not in my exponent, and my exponents here are positive whole numbers so it looks like, yes, this is a polynomial function. You can have a fraction as a coefficient, just not as an exponent. So let's go ahead and write this in standard form, which I can do by simply combining these like terms that I have here. So I end up with f(x)=-23x4+4, combining that 1 and 3. So let's identify our degree here.

So my highest power on that first term is 84. And then lastly, my leading coefficient, looking here, I have negative 2 thirds as my leading coefficient on that term with the highest power. Okay. So we've taken a look at some polynomial functions. Let's look at what their graphs may look like.

So there are 2 things that we want to think about with the graphs of polynomial functions, and that is that they are both smooth and continuous. This will be true of the graph of any polynomial function, and what that means is that there will be no corners in our graph and there will be no breaks. So looking at these polynomial functions on the left side, I have this curve here that is smooth. It is a smooth curve; it is continuous.

It never breaks off. And then here, you might recognize this as a quadratic function, which we know is definitely a polynomial function, and it is both smooth and continuous as well. Now looking over to this right side here and this graph, I have a really harsh corner right here, so that tells me that I am not dealing with a polynomial function. It is not a polynomial function at all. It is not smooth and continuous.

So let's look at one more here and looking at this graph, it breaks off and then it keeps going on that side. Now this is not okay; it is not a polynomial function because it has that break. So these 2 are not polynomial functions at all, and we want to look for things that look more similar to these 2 that are both smooth and continuous. So one more thing that I want to mention about the graphs of polynomial functions is the domain, which is always for any polynomial function going to go from negative infinity to infinity, which is something you may remember from working with quadratic functions. All quadratic functions had this domain and all polynomial functions have it as well.

All real numbers are included in that domain. So that's a little of the basics of polynomial functions. Thanks for watching, and I'll see you in the next video.

Determine if the given function is a polynomial function. If so, write in standard form, then state the degree and leading coefficient. $f\left(x\right)=4x^3+\frac12x^{-1}-2x+1$

• A. Polynomial with $n=3,a_{n}=4$
• B. Polynomial with $n=4,a_{n}=3$
• C. Polynomial with $n=-1,a_{n}=\frac12$
• D. Not a polynomial function.

Determine if the given function is a polynomial function. If so, write in standard form, then state the degree and leading coefficient. $f\left(x\right)=2+x$

• A. Polynomial with $n=1,a_{n}=2$
• B. Polynomial with $n=0,a_{n}=1$
• C. Polynomial with $n=1,a_{n}=1$
• D. Not a polynomial function.

Determine if the given function is a polynomial function. If so, write in standard form, then state the degree and leading coefficient. $f\left(x\right)=3x^2+5x+2$

• A. Polynomial with $n=3,a_{n}=2$
• B. Polynomial with $n=2,a_{n}=3$
• C. Polynomial with $n=2,a_{n}=2$
• D. Not a polynomial function.

Intro to Rational Functions

Hey, everyone. By now we've worked with a bunch of different polynomial functions, things like x2 + 4x + 1 or 3x + 2 or other polynomials of different degrees. But if we take these polynomials and put one on the top of a fraction and the other on the bottom of a fraction, we now have a new function called a rational function that has a polynomial of any degree in both the numerator and denominator of a fraction. You might see this written in your textbooks as p(x)∕q(x), where p(x) just represents any polynomial and q(x) is any other polynomial. I know that fractions might not be your favorite thing to work with, and taking polynomials and putting them into fractions might seem a little bit intimidating.

But we've worked with polynomial functions before, and we've even already worked with rational equations. So, in working with rational functions, we're going to rely a lot on what we already know about these functions and equations that we've already worked with. And I'm going to walk you through everything you need to know step by step. So, let's go ahead and get started.

Now, when working with rational equations, we said our denominator could not be equal to 0. So for this particular equation, looking at my denominator, I would say that x≠1, and that would be referred to as my restriction because 1 would make my denominator 0, which is not something that I want to happen. The same thing is true of rational functions because the same thing is true of all fractions. The denominator can never be 0. Now, because we're working with a function now that can be evaluated for any value of x, we're going to refer to this as our domain restriction.

It's the same exact idea. Our denominator cannot be 0. We're now just referring to our entire domain and defining what x can and can't be based on that restriction. So looking at this function that I have here, 1x-1, I have the same restriction. But writing it in my domain, I write my domain in set notation in x such that x≠1 because that's what would make my denominator 0.

Now we write it like this because we determine our domain the same exact way. We simply set our denominator equal to 0 and solve for x to get that restriction. And then we know that our domain can be any real number besides that number that makes our denominator 0. So, now that we know how to find the domain, let's go ahead and find the domain of a couple of different rational functions here. Now looking at this first rational function, I have 33x-12.

We're simply going to take our denominator and set it equal to 0, solving for x. So subtracting 12 from both sides, I end up with 3x=-12. And to isolate x, I go ahead and divide both sides by 3. That leaves me with x=-4, which tells me my domain restriction. I know that x can be any real number, except x≠-4 because if I took negative 4 and plugged it back into my denominator, it would make it 0, which is what I don't want.

So my domain here is x such that x≠-4. Now let's look at another function here. So I have f(x)=x+5x2-25. Again, taking our denominator, setting it equal to 0, if I add 25 to both sides, which cancels out that 25, leaving me with x2=25. And to isolate x, I simply take the square root of both sides, leaving me with x=+or-25, which I know is just 5.

Now with this answer, this tells me my domain is restricted by 2 values. It can't be 5 or -5. So my domain is x such that x≠5 or x≠-5. We can have multiple domain restrictions depending on what our denominator is. Now that we know how to find the domain of different functions, something else youll commonly be asked to do when working with rational functions is to write them in lowest terms.

Now, in order to write functions in lowest terms, we're going to factor something we've done a million times before. We're going to factor both the top and bottom of our function and then simply cancel out any common factors. So we're just writing our function in the simplest terms, simplifying it as fully as we can. So let's take another look at these functions here. So starting again with 33x+12, let's go ahead and factor the top and the bottom here.

I can't really factor the top because it's just the constant. So I'm simply left with 3 on the top there. But on the bottom, it looks like I can pull out a greatest common factor of 3 from both of these terms, leaving me with 3×x+4. Now we want to go ahead and cancel out any common factors. In this case, I have a common factor in my numerator and my denominator of 3, so I can go ahead and cancel that out, leaving me with my function 1x+4 as all thats left.

My function in lowest terms here is simply 1x+4. Now let's look at our other function. So we have f(x)=x+5x2-25. So let's go ahead and factor. Now again, my numerator can't be factored, it's already just x+5, but my denominator is a difference of squares so this factors to x+5×x-5.

Now I want to go ahead and cancel our common factors. In this case, I have a common factor of x+5 that I can cancel from both my numerator and my denominator. So canceling that factor out leaves me with 1x-5 and that's my function in lowest terms. Now something I want to point out here is that if I were to take this function in lowest terms and solve for the denominator here by setting my denominator equal to 0, I would only get one of the values that appeared in my domain restriction up here. I would say that x≠5, but I wouldn't know about that negative 5.

That's because you always need to find your domain before you write in lowest terms. Never write in lowest terms and then find the domain because you might be leaving something out. Okay, now that we know the basics of working with rational functions, let's move on. I'll see you in the next video.

Find the domain of the rational function. Then, write it in lowest terms.

$f\left(x\right)=\frac{x^2+9}{x-3}$

• A. {x∣x≠0}, $f\left(x\right)=\frac{1}{x-3}$
• B. {x∣x≠3}, $f\left(x\right)=\frac{x^2+9}{x-3}$
• C. {x∣x≠−3}, $f\left(x\right)=\frac{x^2+9}{x-3}$
• D. {x∣x≠3}, $f\left(x\right)=x+3$

Find the domain of the rational function. Then, write it in lowest terms.

$f\left(x\right)=\frac{6x^5}{2x^2-8}$

• A. ${\left\lbrace x|x\ne2,-2\right\rbrace},f(x)=\frac{3x^5}{x^2-4}$
• B. ${\left\lbrace x|x\ne2,-2\right\rbrace},f(x)=\frac{6x^5}{2x^2-8}$
• C. ${\left\lbrace x|x\ne2\right\rbrace},f(x)=\frac{3x^5}{x^2-4}$
• D. ${\left\lbrace x|x\ne2\right\rbrace},f(x)=\frac{3x^5}{x^2-8}$

Exponential Functions

Welcome back, everyone. By now, we've worked with a ton of different polynomial functions, and one of the most common and perhaps most basic polynomial functions we've seen is \( f(x) = x^2 \). But if I take that \( x \) and that \( 2 \) and I swap them, I now have \( f(x) = 2^x \), which is an entirely different type of function called an exponential function that we're going to have to evaluate and graph just as we have other functions. Now, I know seeing that variable in the exponent might seem a little bit strange at first, but here I'm going to walk you through the basics of exponential functions using a lot of what we already know about exponents themselves and what we know from working with other functions. So let's go ahead and get started.

Now looking at our function here, \( 2^x \), we have 2 different things going on. We have this base number of 2, and then we have the power that that base 2 is raised to, in this case, \( x \). Now, in working with exponential functions, we need to consider a couple of different things about the base of our function. So in this case, we have 2. But the base of any exponential function needs to be a constant, so it can't be something with a variable that changes.

It has to be positive. So we can't have any negative numbers as our base. And it also cannot be equal to 1. So something like 2 fits all of that criteria. Now when considering the exponent or the power of our exponential function, we only need to consider one thing, and that is that it contains a variable.

So it's something that can change, something like \( x \). Now that we know a little bit about exponential functions, let's take a look at a couple of different exponential functions down here. So here we have this function \( f(x) = \left(\frac{2}{3}\right)^x \). And here we want to determine if this even is an exponential function. And if it is, then we can go ahead and identify both the power and the base.

So looking at this function here, I have \( \frac{2}{3} \) as that base. And I want to consider my three things constant positive and not equal to 1. So \( \frac{2}{3} \), it is constant. It is positive. And it is not 1.

So it looks like we're good so far. Now looking at our power here, we have this power of \( x \). We need to make sure our power contains a variable, which it already is a variable just by itself. So it looks like we are good here, and this is an exponential function. Now to identify our power and our base, we saw our power of \( x \) there, so our power is simply \( x \) by itself.

And then our base is \( \frac{2}{3} \). Now it's okay to have a base that's a fraction as long as it's constant and positive. Let's move on to our next example. So here we have \( f(y) = 1^y \). Now our base here is 1.

1 is constant. It is positive, but it is 1, so it is not meeting that last criteria not being equal to 1. So I don't even have to look any further. I already know that this is not an exponential function. I don't need to worry about identifying my power and my base.

\( 1^y \) is just going to be 1, so that power of \( y \) doesn't even matter, which is why this isn't an exponential function at all. Let's look at one final example here. Here we have \( f(x) = 10^{x+1} \). And we want to identify if this is an exponential function. So here my base is 10.

So this 10 is constant. It is positive. And it is not 1. So we're looking good so far. Now looking at our power here of \( x+1 \), it does contain a variable.

It has that \( x \). So it looks like, yes, we are dealing with an exponential function here. Now here, our power is this entire thing, \( x+1 \). It's not just the variable by itself. It's everything that our base is being raised to.

Now here, our base is this 10 because that's what's being raised to the power of \( x+1 \). So we are good here with our power of \( x+1 \) and our base of 10. Now that we know how to identify an exponential function, let's go ahead and get into evaluating them. Now we're going to have to evaluate exponential functions for different values of \( x \), which just means we're going to plug in values of \( x \) to our function. So here we have \( f(x) = 2^x \).

So to evaluate this for \( x = 4 \), I'm simply going to plug 4 in for \( x \). So here, \( f(4) \) is really just \( 2^4 \). Now, using my rules for exponents, I know that \( 2^4 \) is really just 2 times 2 times 2 times 2 4 times, which will give me 16 as my final answer. Let's move on to our next example here and evaluate our function for \( x = -3 \). Now, plugging 3 in for \( x \) here, \( f(-3) \), gives me \( 2^{-3} \) because we're still using that same function here.

Now whenever I have a negative in the exponent, that's totally fine. It just means that I'm really dealing with \( \frac{1}{2^3} \) because it's really a fraction. Now with this \( 2^3 \) on the bottom, this is really just 2 times 2 times 2, 3 times. So my answer here is simply \( \frac{1}{8} \). I end up with this fraction.

Now let's move on to our next example. We have \( x = 3.14 \). So, of course, plugging that 3.14 in for \( x \) to our function, we have \( f(3.14) = 2^{3.14} \). Now this is not something that I want to or I may have been capable of doing by hand, so this is when I would actually want to use my calculator. So if you're ever unsure of how to do it by hand, just type it into your calculator.

Now the first thing we're going to do is type in our base, in this case, 2. Now in order to raise something to a power to get that exponent, we're going to use the caret key that looks like this on your calculator. So I'm going to take 2, raise it to the power using that caret key, and then I'm simply going to type my power in, in this case, 3.14. Now you can always add some parentheses around everything if you just want to make sure that everything is going into your calculator the way that you need it to. So if I type this in my calculator, I take 2 and raise it to the power of 3.14, I end up with about 8.815 as my final answer.

Now let's take a look at one final example here. Here we have \( x = 12 \). So in order to plug 12 into my function for \( x \), I can go ahead and take \( f(12) \), and that gives me \( 2^{12} \). Now I don't really want to multiply 2 by itself 12 times. So if your exponent is rather large and you don't want to do it by hand, go ahead and type that into your calculator.

So here we would do \( 2 \text{ caret } 12 \). And in our calculator, we would end up getting 4,096 as our final answer. Now that we have a good idea of what to do when working with exponential functions and how to evaluate them, thanks so much for watching. And I'll see you in the next one.

Determine if the function is an exponential function.  

If so, identify the power & base, then evaluate for $x=4$.

$f\left(x\right)=\left(-2\right)^{x}$

• A. Exponential function, $f\left(4\right)=16$
• B. Exponential function, $f\left(4\right)=-16$
• C. Not an exponential function

Determine if the function is an exponential function.  

If so, identify the power & base, then evaluate for $x=4$ .

$f\left(x\right)=3\left(1.5\right)^{x}$

• A. Exponential function, $f\left(4\right)=410.06$
• B. Exponential function, $f\left(4\right)=15.19$
• C. Not an exponential function

Determine if the function is an exponential function.  

If so, identify the power & base, then evaluate for $x=4$ .

$f\left(x\right)=\left(\frac12\right)^{x}$

• A. Exponential function, $f\left(4\right)=\frac{1}{16}$
• B. Exponential function, $f\left(4\right)=-16$
• C. Not an exponential function

Logarithms Introduction

Hey, everyone. Whenever we solve polynomial equations like \( x^3 = 216 \), we can simply do the reverse operation in order to isolate \( x \). So here, since \( x \) is being cubed, I could simply take the cube root of both sides in order to cancel that, isolating \( x \), giving me my answer that \( x \) is equal to the cube root of 216. But what if my variable \( x \) is instead in my exponent, like this \( 2^x = 8 \)? How are we going to isolate \( x \) here?

Well, I know that looking at this, I can just think, okay, how many times does 2 need to be multiplied by itself in order to give me 8? And I know that my answer is simply \( x = 3 \). But what if I'm given something like \( 2^x = 216 \)? I really don't want to have to multiply 2 enough times in order to get up to that 216. So is there not just an operation I can do in order to cancel out that 2 and leave me with \( x \)?

Well, here I'm going to show you that there is an operation that does just that because the reverse or inverse of an exponential is actually taking the logarithm or a log. Now, the first time you work with logs, they can be a little bit overwhelming. But here I'm going to walk you through exactly what a log is and how we can use it to actually make our lives easier, especially when working with exponents. So now that we know that the reverse or inverse operation of an exponential is simply taking the log, we can go ahead and take the log of both sides of this equation in order to isolate \( x \). So I take log of \( 2^x \), and that's equal to log of 216.

But we actually do need to consider one more thing here because whenever we canceled the 3 on our \( x^3 \), we took the cube root. We didn't take the square root or the 4th root or anything else. We took the cube root in order to cancel that 3. And we need to consider something similar when working with logs because logs and exponentials need to have the same base as each other in order to cancel. So here, since I have an exponential of base 2, I want my log to have a base of 2 as well.

So really, I want to take a log base 2 of \( 2^x \) and then on the other side, a log base 2 of 216. Now that my log and my exponential have the same base, then it's going to fully cancel out, leaving me with just \( x = \log_2(216) \). Now it's fine to leave it in this form here. This is actually called our logarithmic form. And we're later going to learn how to fully evaluate these and get a number, but for now, we're just going to keep it in that log form.

Now that we're here, what exactly does this statement mean? We have \( \log_2(216) \). Well, a log is actually giving us the power that some base must be raised to in order to equal a particular number. But what does all of that mean? Well, looking at our function here or our equation here, \( \log_2(216) \), this is really saying, okay, what power does 2 need to be raised to in order to give me 216?

Now, this statement here I mentioned is in its logarithmic form, and it's actually an equivalent statement to our very first equation, \( 2^x = 216 \). This is just in its exponential form, and we basically translated it into its logarithmic form. Now we're going to have to do this for multiple statements. We're going to have to translate and convert expressions between these two forms. So diving a bit deeper in converting between these two forms, let's start by taking this equation in its exponential form, \( 3^x = 81 \), and putting it into log form.

Now whenever we convert between these two forms, no matter what we're going to or from, we're always going to start at the same place. We're going to start with our base. So this base 3 of our exponent is going to become the base of our log. So I start here with \( \log_3 \). Now once I have my base, I'm going to circle to the other side of my equal sign to that 81.

So I have \( \log_3(81) \), and then once I have that 81, I'm going to circle back to the other side of my equal sign, and that's what \( \log_3(81) \) is going to equal, so that's equal to \( x \). Now I have my equivalent statement in its log form. \( \log_3(81) = x \) is equivalent to \( 3^x = 81 \). Now that we've seen going from exponential to log form, let's go in the reverse direction from log to exponential form. So starting with this statement here, \( x = \log_4(64) \), we're going to start at that same place.

We're going to start with our base. So here, my log has a base of 4. That becomes the base of my exponential. So I have a log base 4, and I have that 4, and now I want to raise 4 to a power, and I'm going to raise it to the power that is on the other side of my equal sign. So start with your base, go to the other side of your equal sign.

In this case, I get \( x \) here, so \( 4^x \), and then finally circling back to the other side of our equal sign, that is our final number here, 64. So \( 4^x = 64 \) is an equivalent statement in exponential form from this \( x = \log_4(64) \). Now I know that that might seem like a lot right now, so let's just walk through some examples together. So let's start with this \( x = \log_5(800) \). Since that is in its log form, I have a log right there, I wanna go ahead and put this in exponential form.

So remember, we're going to start at the same place every single time no matter what we're starting with. We always start with that base. So here I have log base 5, so I'm going to start with an exponential of base 5. Now once I have that base, I'm going to circle to the other side of my equal sign and get that \( x \). So \( 5^x \).

And then I'm gonna circle back to the other side of my equal sign, and this is equal to 800. So start with your base, other side of your equal sign, circle back to where you started. So \( 5^x = 800 \) is my equivalent statement in its exponential form. Let's look at another example here. We have \( \log_2(16) = 4 \).

Now where are we going to start here? Well, we wanna start with our base, of course, so I have this log base 2 so that becomes the base of my exponent. So I have 2 and then I go to the other side of my equal sign \( 2^4 \) and then circle back to where you started. \( 2^4 = 16 \). Now this is great because we can actually see that this is a true statement.

\( 2^4 = 16 \), so I know that I wrote that correctly. Let's look at one final example here. Here we have \( 10^x = 45100 \). So this is in its exponential form. We wanna go ahead and put this in log form.

So remember, we're going to start with our base. So here, my exponent has a base of 10, and this becomes the base of my log. So I start with \( \log_{10} \), and then I circle to the other side of my equal sign and get 45100. So \( \log_{10}(45100) \) and that is equal to circling back to the side that I started on, \( \log_{10}(45100) = x \) and that is my equivalent statement. But looking at this \( \log_{10}(45100) \), this log base 10 is actually a sort of special type of log.

So log base 10 is actually known as the common log because it occurs so frequently. So it can actually just be written as log. It gets its own special notation. It's just log because it is that common. So I could really just write this as a log of 45100, and that is equal to \( x \).

Now this also has its own button on your calculator if you need to evaluate, which we'll do in the future. It's just the log button. So now that we know a little bit more about logs and we've even seen our first common log, let's go ahead and get some more practice.

Graphs of Logarithmic Functions

Hey, everyone. Earlier when graphing exponential functions, we found and plotted a bunch of different ordered pairs, and we found that our graph had similarities to both polynomial and rational functions as we approached an asymptote. Now we're faced with graphing another type of function, a logarithmic function, and you may be dreading having to learn how to graph yet another different type of function. But I have great news for you here because graphing a log function is actually super similar to graphing an exponential function because a log is the inverse of an exponential. So here, I'm going to walk you through graphing a log function using exclusively things we already know about graphing exponential functions.

So let's go ahead and get started. Now looking at the log function we have here, we have a log base 2 of x. So let's start by just plugging in some values of x and getting some ordered pairs. So starting with x equals 1, if I plug that into my function, I get log base 2 of 1. And I know any time I take the log of 1, no matter what the base is, I'm always going to get 0.

So I have my first ordered pair at (1,0), which I can go ahead and plot on my graph. Then if I plug in x equals 2 to my function, I get log base 2 of 2, which since my base is the exact same as what I'm taking the log of, I know that my answer here is going to be 1, and I get my 2nd ordered pair at (2,1). Now let's pause here and take a closer look at these two points. I have a (1,0) and (2,1). Now looking over to my exponential function here, I have f of x is equal to 2^x, which is the inverse function of this log function that we are trying to graph here.

Now looking at my exponential function, I have these two points, (0,1) and (1,2). And looking back to the points that we just found, I noticed that these are the same exact points with x and y flipped. Now this is because these are inverse functions, so this is actually going to work for every single x and y value of our exponential function. If we simply flip them, we'll have all of the new ordered pairs needed to graph our log function. So let's go ahead and flip our x and y values and get these points plotted on our graph.

So looking at our first point here, flipping (0,1), we have my new point, (1,0). And then for the (1,2) point, I know f(x) is 2, so flipping these final two points here, I'll end up with (2,1). And we can go ahead and plot all of these on our graph. So (2,1), and for our small fractions, (1/2, -1) and (1/4, -2).

Now connecting all of these points, I have a complete graph of my log function. I notice here on this side that we're getting really close to that y axis but not quite touching it, which tells us that we're, of course, dealing with an asymptote that I can go ahead and plot using a dashed line right along that y axis to form my vertical asymptote. Now that we have the complete picture of the graph of our log function, let's compare it to our original exponential function. Now looking at the graph of my exponential function here, if I were to take my entire graph and fold it along this diagonal line and then stamp my exponential function onto the other side, I would end up with a graph identical to the log function that we just graphed because this is actually exactly what we did. We simply reflected the graph of our exponential function over this diagonal line, y equals x, in order to get the graph of our log function.

Now, this will work for any log function and its corresponding inverse exponential function because they're inverses. They're always going to be reflections of each other along the line y equals x. Now, if you want a way to remember the shapes of these graphs, remember when you're dealing with an exponential function, you can think of a lowercase 'e'. And if we just extend the tail of that 'e' out, it looks really similar to the shape of the graph of our exponential function. And for a logarithmic function, if we look at our word logarithmic and we take that lowercase 'r' and extend that 'r' out, we get a shape really similar to the graph of our logarithmic function.

So remember, for an exponential function graph, the shape looks really similar to extending a lowercase 'e', whereas for a logarithmic function, we extend that lowercase 'r'. Now that we have a good idea of what's going on with the shapes of these graphs and what they look like, let's also consider their domain and range. Now when graphing our log function, we were able to flip the x and y values of the inverse exponential function, and we can actually do the same exact thing to our domain and range. So the domain of our exponential function corresponds to the range of our inverse log function, which is going to be all real numbers for any log function of any base.

Then, also, the range of our exponential function corresponds to the domain of our log function, and it's going to depend on our asymptote and in this case, go from 0 to infinity. Now speaking of asymptotes, we also can take a look at our asymptote here, and we had a horizontal asymptote at y equals 0 for our exponential function, which then corresponds to a vertical asymptote now at x equals 0 for our logarithmic function. Now let's consider one final thing here and consider when we have to graph logs of different bases. Now, everything that we've looked at so far has been kind of the opposite of working with an exponential function but this thing is going to be the exact same and the direction of our graph is going to depend on the value of our base b just as it did for exponential functions and it's going to depend on it in the exact same way. So for values of b that are greater than 1 like 2 or 3 or 10 or whatever, our graph is going to be increasing.

And for values of b between 0 and 1, like, say, 1/2, our graph is instead going to be decreasing. Now that we know everything that we need to about graphing log functions, let's get some more practice.

Graphing Logarithmic Functions Example 1

Hey everyone. In this problem, we're asked to graph 2 different functions, \( f(x) = 3^x \) and \( g(x) = \log_3(x) \). Now we can see that these functions are inverses of each other because the logarithm is the inverse function of an exponential, and we're going to go ahead and graph both of them together. So let's start with \( f(x) = 3^x \) and find and plot some points here. Starting with \( x = 0 \), we can plug that into our function and get \( 3^0 \), which is just 1.

Now, if I plug 1 into my function, that's \( 3^1 \), which is just 3. And then \( 3^2 \) is going to give me 9. Now, let's go ahead and plot these points on our graph. We know that it's going to increase really rapidly and get really steep there. Now for my negative values of \( x \), we're going to see the same numbers but in fractions as with positive \( x \). So \( 3^{-1} \) is simply \( \frac{1}{3} \) and then \( 3^{-2} \) is \( \frac{1}{9} \) and we can see that we're getting smaller and smaller as we go into those negative numbers. Let's go ahead and plot those points at negative one, \( \frac{1}{3} \), and then 2, \( \frac{1}{9} \), really close to my x-axis. We can go ahead and connect all of these points that we have on our graph here.

I know that I'm going to get really steep on this side. And then on my left side here, I'm going to get really close to my x-axis but not quite cross it because we have an asymptote there. So I can go ahead and plot my asymptote as well. I know that it's at the x-axis right at \( y = 0 \). Now that we have our graph for \( 3^x \), let's move on to plotting \( g(x) = \log_3(x) \).

Now, remember that whenever we're dealing with inverse functions, we can simply swap our \( x \) and \( y \) values. So these values that I have for \( x \) over here are simply going to become my \( y \) values when working with my new function that is the inverse. And then my \( y \) values over here are going to become the \( x \) values of my inverse function. Let's go ahead and swap all of those points so that we can get them plotted on our graph. Starting with this point, it now becomes \( (1, 0) \). Then I have \( (1, 3) \) which is flipped to get \( (3, 1) \). And then my last point, flipping that 2 and that 9, I get \( (9, 2) \).

Now let's go ahead and plot all of these points for our function \( g(x) \) to get our final function up here. Starting with \( (1, 0) \) and working our way down this way first. Then \( (3, 1) \). And \( (9, 2) \) which goes off my graph yet again. And then going the opposite direction into my fractions, \( \frac{1}{3} \) at \( -1 \) is going to be right about here. And then \( \frac{1}{9} \) at \( -2 \), we're going to be getting really close to our \( y \)-axis this time. So we can go ahead and connect all of these points to form our graph. And then we're getting really close to that \( y \)-axis, which tells us that we're dealing with a vertical asymptote up on this side right on our \( y \)-axis at \( x = 0 \). So now that we have plotted both of our functions, we can go ahead and determine our domain and range.

Now when managing our domain of our first function, \( f(x) = 3^x \), I can see that this is going to be any value of \( x \) - it can be all real numbers. Naturally, our domain of \( 3^x \) will match up with the range of our function \( \log_3(x) \). The range of \( 3^x \) starts from my asymptote at point 0, all the way to infinity \( (0, \infty) \). Conversely, this range flips to become the domain of our other function, the inverse. This indicates that the domain of \( \log_3(x) \) is going to be from 0 to infinity, which we can again verify by examining our graph. We see that we cannot cross this asymptote, so we start at the 0 point and extend to everything positive.

Now that we've graphed these inverse functions, let's get some more practice. Let me know if you have any questions.