Curve Sketching - Online Tutor, Practice Problems & Exam Prep

Summary of Curve Sketching

Throughout this chapter, we've learned a ton of information about a function's first and second derivative and how that gives us information about a function's graph, like where it's increasing or decreasing, where it's concave up or down, and even the locations of local maxima and local minima. So we can use all of that information along with our basic knowledge of graphing to actually sketch the graph of a function, and that's exactly what we're going to do here. Now this can be a bit of a tedious and time-consuming process, but you already have all of the knowledge that you need to be able to graph a function based on its first and second derivative. And I'm going to be here to walk you through this step by step. So let's go ahead and get started.

Now here, we're tasked with sketching the graph of f(x)=x3−3x2+4. Now our first four steps here are just going to focus on the basics of graphing before we get into our first and second derivative. Now starting with our domain here, the domain of our function, because this is a basic polynomial function, I know that my domain is all real numbers. I have no domain restrictions, so my domain goes from negative infinity to positive infinity. Now, for our x and y intercepts here, remember that in order to determine the x intercepts of a function, I just need to set that function equal to 0.

Now when working with a polynomial function such as this one, it's going to be helpful at this point to have your function factored. Now if you need a refresher on factoring, feel free to look back to earlier videos. Now for this factored polynomial function, we end up with x−2x+1. Setting this equal to 0 means that I can set each of these individual factors equal to 0. So starting with that x−2, if I set that equal to 0 and add 2 to both sides, I get x=2.

Then if I take x−1, set that equal to 0, this will result in x=−1. So I have my 2 x intercepts, and I can go ahead and plot them on my graph over here. x=2, and x=−1. Now for my y intercept, we just want to plug 0 into our function and solve. These first two terms end up going away.

They're just 0. So we're just left with a y intercept of 4, which, again, I can go ahead and plot on my graph here on my y axis. So I have all of my intercepts there. Let's move on to step 3, finding any asymptotes. Now because again this is a basic polynomial function, it does not have any asymptotes.

We don't have to worry about this step. We can move on to step 4, determining the symmetry of our function. Now knowing information about a function's symmetry can be helpful because it just gives us more guidance when graphing. In order to determine the symmetry of a function, we want to plug negative x in. Now doing that here, this ends up simplifying to −x3−3x2+4.

Comparing this to our original function, it is definitely not the exact same function because I have this negative here. So that doesn't tell me that my function is symmetric about the y-axis. It is also not the fully negative version of my function. So we don't know that it's symmetric about the origin either. And this didn't give us any information about this function's symmetry.

But with all of this out of the way, we can go ahead and move on to using our first derivative here. Remember, the sign of the first derivative of a function tells us whether that function is increasing or decreasing. We can determine these intervals by first finding our critical points. This is where that first derivative is either equal to 0 or does not exist. So differentiating that original function using the power rule, we end up getting a 3x2−6x, which factors to 3x∗x−2.

Setting each of these individual factors equal to 0 results in 2 critical points, one where x=0 and one where x=2. There's nowhere that this first derivative does not exist, so these are our only 2 critical points. We can go ahead and create our intervals on our sign chart here by marking our 2 critical points, 0 and 2. This splits our sign chart into 3 intervals, from negative infinity to 0, from 0 to 2, and from 2 on to infinity. Now, we can determine where our function is increasing or decreasing on each of these intervals by choosing a test value in each interval to plug back into that derivative.

Remember, we want to choose test values that are just going to make it easy for us to work with. So on this first interval from negative infinity to 0, I'm going to choose a negative one. Then from 0 to 2, I'll choose a positive one. And finally, from 2 on to infinity, I'll choose positive 3. Now feel free to pause here and plug these values back into that first derivative on your own before checking back in with me.

Remember, at this point, we only care about the sign of these values. That will tell us whether our function is increasing or decreasing. So when you plug a negative one into that first derivative, you should get a positive value. Then if you plug positive one in, you should get a negative value. Finally, if you plug 3 into that first derivative, you should again get a positive value.

So what does that tell us about where our function is increasing or decreasing? Well, on that first interval from negative infinity to 0, that tells us our function is increasing, that positive derivative. I'm going to go ahead and mark this on my function here. From negative infinity to 0, our function will be increasing. Then on this second interval from 0 to 2, it has a negative derivative.

That tells us there, our function will be decreasing. Then finally, on that last interval from 2 onto infinity, that positive derivative tells us that our function will be increasing there. So we now know where our function is increasing or decreasing. But what about its concavity? This will tell us exactly how to sketch the shape of our graph.

Moving on to concavity here, remember that the sign of our second derivative tells us whether a function is concave up or concave down. Now we want to start here by finding any potential inflection points. That's where that second derivative is either equal to 0 or does not exist. Now differentiating that first derivative again using the power rule, we end up getting a second derivative of 6x−6 or factored to 6∗x−1. If I set this equal to 0, I get one potential inflection point at x=1.

There's nowhere that the second derivative won't exist. So this is my only potential inflection point that I can go ahead and mark on my sine chart here. This splits my function into 2 intervals from negative infinity to positive one and from 1 on to infinity. Now the same way we choose test values when using our first derivative, we want to choose some test values here. Based on what's already on my sign chart, I'm going to go ahead and choose a test value of 0 and a test value of 2 to plug back into that second derivative.

Remember that we only care about the sign here when we plug these values in. Plugging 0 in should result in a negative second derivative. Plugging 2 in results in a positive second derivative. So what does this tell us about our function's concavity? Well, on that first interval from negative infinity to positive one, that tells us our function is concave down there.

It has a negative second derivative. So I'm going to go ahead and mark that on my graph here concave down from negative infinity to 1. Then on that second interval from 1 on to infinity, that positive second derivative tells me that my function is concave up. So marking that again on my graph here Based on the fact that my function changes from being concave down to concave up, this means that x=1 is indeed an inflection point. It can be helpful to go ahead and plot your inflection point so that you know exactly where the concavity changes.

If we plug a 1 back into our original function, this results in an ordered pair at 12, which I can go ahead and plot up here on my graph so that I know exactly where my function changes from being concave down to being concave up. So I now have a ton of information from my first and second derivative. Now because of what we have learned about our first and second derivative, we know that this now allows us to determine where our local extremum are located. Remember that you could use the first or second derivative test to do this. Here, I'm going to use the first derivative test, which focuses on where the sign of our first derivative changes.

So going back to what we did with our first derivative here, we can see that at these critical points, x=0 and x=2, that the sign of that first derivative does change. At x=0, my function goes from having a positive derivative to having a negative derivative, increasing to decreasing. That tells us, based on that sign change, that this point is a local maximum. So we have a local maximum at x=0. Then at x=2, my function goes from having a negative derivative to then having a positive one.

It goes from decreasing to increasing. That tells us that this point is a local minimum based on that sign change. So we have a local minimum at x=2. Now the actual location of these two values, remember, we can find them by plugging them back into our original function. It turns out that these are values that we have already plotted on our graph because there are x and y intercepts.

F(0) was 4 and F(2) is 0. These are points that we already have here. But knowing now that they are local max and local min, we can indicate that on our graph as well. We know that this point at x=0 is a local max, so it's going to have some sort of peak there. And then at x=2, this is a local min.

It will have some sort of valley. So now we've found a ton of information about our function. And the only thing that we have to do now is connect all of the points that we've already plotted with a smooth curve based on everything that we know about our function from steps 4 through 6. So we know where our function is increasing or decreasing, where it's concave down and concave up. We now just need to sketch our graph based on that knowledge.

So taking a look at our graph from left to right, we can see on this far left side that our function starts off increasing and concave down. Remember, concave down looks like a frown, and increasing means that we're going to be going up. So because our function starts off increasing and concave down, I'm going to start my graph down here knowing that it has that x intercept right here. Then it has this local maximum at x equals 0, so increasing up to that point. I'm going to sketch my function here.

It is concave down and increasing. At that point, x equals 0, my function begins to decrease. It is still concave down at this point, so concave down looking like a frown. And then it continues decreasing and being concave down until we reach that point of inflection. At this point, my function continues decreasing, but it is now concave up looking like a smile.

So continuing to decrease but changing that curve a little bit, we come back down to this x equals 2. This is a local minimum. So I know that I'm going to have some sort of valley there. We know that at this point, my function is still concave up, but it is now increasing. And this remains for the rest of my function.

So I can see here that I have now fully graphed my function based on everything that we found. So when faced with sketching the graph of a function, start with your basics of graphing: the domain, your intercepts, your asymptotes, and the symmetry. Then you can use your first and second derivative to determine where your function is increasing or decreasing, where it's concave up or down, and where your local extrema are located. Once you're able to sketch your graph based on all of that information, you are good to go. Now, again, I know that this is a long and tedious process, but you got this.

We're going to practice some more together coming up next.

Summary of Curve Sketching Example 1

In this problem, we're asked to sketch the graph of the rational function \( f(x) = \frac{x - 1}{x + 2} \). We'll use our basic knowledge of graphing along with what we know about the first and second derivatives of a function to sketch this graph. Sketching the graph of a rational function can feel a bit more complicated because there's more going on compared to dealing with a polynomial, but we'll still follow the same steps. Let's jump into step 1 here and determine the domain of our function. Since we're working with a rational function, our denominator cannot be equal to 0.

Establishing our domain, we know that \(x + 2\) cannot be equal to 0. Subtracting 2 from both sides, we find that \(x\) cannot be equal to -2, which is our domain restriction. So, our domain extends from negative infinity to -2 (exclusive), and from -2 to infinity. Next, we move to step 2 to find our x and y intercepts. We start by setting our function equal to 0, thus setting the numerator \(x - 1 = 0\). Adding 1 to both sides gives \(x = 1\), which is our x-intercept.

Then, plugging 0 into our function, \(f(0) = \frac{0 - 1}{0 + 2} = -\frac{1}{2}\), gives us our y-intercept. Now, we move on to step 3 to find asymptotes. For a vertical asymptote, we set the denominator equal to 0, resulting in \(x = -2\), indicating a vertical asymptote at \(x = -2\). For horizontal asymptotes, we compare the degrees of the numerator and the denominator, which are the same. The leading coefficients are both 1. Hence, we have a horizontal asymptote at \(y = 1\).

In step 4, we check the symmetry by plugging \(-x\) into the function, resulting in \(\frac{-x - 1}{-x + 2}\). Factoring out a negative, we get \(\frac{x + 1}{x - 2}\), which is neither the original function nor its negative, so no symmetry information is inferred. Next, we use the first derivative for step 5 to determine where the function is increasing or decreasing. Applying the quotient rule, our first derivative is \(f'(x) = \frac{(x + 2) \cdot 0 - (x - 1) \cdot 1}{(x + 2)^2} = \frac{-x + 1}{(x + 2)^2}\).

Simplifying, we find \(f'(x) = \frac{3}{(x + 2)^2}\), indicating that the function is always increasing as the derivative is always positive. For step 6, we find the second derivative to analyze concavity. After differentiation, \(f''(x) = -\frac{6}{(x + 2)^3}\) tells us where the function is concave up and down. Testing values, we find the function is concave up on \((- \infty, -2)\) and concave down on \((-2, +\infty)\).

Finally, as we look at our function in total, we plot additional points to ensure accuracy in our graph, such as \(f(-3)\) and \(f(-4)\). We conclude that the function starts increasing and is concave up to \(x = -2\), then continues increasing but shifts to being concave down after crossing the asymptote at \(x = -2\). Completing this sketch provides a comprehensive view of the behavior of the function \(f(x) = \frac{x - 1}{x + 2}\), illustrating the utility of understanding derivatives and the function’s essential characteristics in curve sketching.

Summary of Curve Sketching Example 2

In this problem, we're asked to graph the given function \( g(x) = \sqrt{x^3 + 8} \). So let's go ahead and just jump right into our steps here. The first thing we want to do is determine the domain of our function. Now looking at this, since it is under a square root, I know that this means that what's under the square root has to be greater than or equal to 0. So \( x^3 + 8 \) must be greater than or equal to 0.

Now if we actually isolate \( x \) here, this then tells us that \( x \) should be greater than or equal to negative 2. This tells us that our domain goes from negative 2, that's an endpoint, on to infinity, and negative 2 is included in that domain. So with that domain out of the way, let's move on to finding our \( x \) and \( y \) intercepts. Remember that in order to find our \( x \) intercepts, we need to set our function equal to 0. To find our \( y \) intercepts, we just need to plug 0 in.

So in this second step here, starting with our \( x \) intercepts, setting \( \sqrt{x^3 + 8} \) equal to 0 here, then we actually already basically solved the same thing here except now we have an equal sign. This will result in the same thing except it's equal, so \( x = -2 \). We can go ahead and mark this \( x \) intercept on our graph right here. Now for our \( y \) intercept plugging 0 in, that's \( 0^3 + 8 \), this results in \( \sqrt{8} \), simplified to \( 2\sqrt{2} \) or as a decimal because this might make it a bit easier to graph. This ends up being about 2.83. So we can go ahead and plot this on our \( y \) axis here. It ends up being right about here.

So with that second step done, we have our intercepts. Now we can move on to step 3 and find any asymptotes. Now since this is a square root function, we don't have to worry about our asymptotes. So there are none of these and we can move on to step 4 in determining the symmetry of our function. Remember, to determine symmetry, we just need to plug negative \( x \) into our function, so completing this step here.

Plugging negative \( x \) in, so we're taking a \( g(-x) \). This is going to be the square root of \( -x^3 + 8 \), which if I simplify this, this just results in \( -x^3 \). So taking that out of the parenthesis because it will still remain negative, plus 8. This, comparing it to our original function, it is not the same thing as our original function so that doesn't tell us any symmetry there and it is also not just the negative version of our original function. So we actually don't gain any new information about our function's symmetry here.

But let's go ahead and shift gears to now using our function's derivatives. So let's go ahead and start with that first derivative, starting here by finding our first derivative. So this is step 5a. We're going to determine where our function is increasing or decreasing. Now remember that our original function \( f(x) \) is \( \sqrt{x^3 + 8} \).

It can be helpful here to rewrite this as \( (x^3 + 8)^{1/2} \) just so that we can easily see how we're going to use the power rule. So finding our derivative here, \( f'(x) \), using the power rule, this becomes \( \frac{1}{2} \times (x^3 + 8)^{-1/2} \). And then I need to multiply this using the chain rule by the derivative of \( x^3 + 8 \), which is \( 3x^2 \). Now if I rewrite this, this becomes \( \frac{3x^2}{2 \times \sqrt{x^3 + 8}} \). Now from here, we want to find our critical points.

This is where this first derivative is either equal to 0 or does not exist. So to find where this is equal to 0, we just want to take that numerator and set that equal to 0. This results in one critical point at \( x = 0 \). Then, in order to find where this derivative does not exist, we can instead take that denominator and set that equal to 0. Now setting that equal to 0, we have already solved this before because this 2 will cancel out and we're just left with \( \sqrt{x^3 + 8} = 0 \).

This results in a critical point at \( x = -2 \). So we have our 2 critical points at \( x = -2 \) and \( x = 0 \). We can go ahead and mark those on our sign chart here. Now looking at this, you may think that it splits our sign chart into 3 intervals, but remember the actual domain of our function. Our function does not exist on this interval from negative infinity to negative 2, so we don't have to worry about that side, and we only have 2 intervals from negative 2 to 0 and from 0 on to infinity.

So now we want to choose a test value in each of these intervals in order to determine where our function is increasing or decreasing. So let's go ahead and choose a test value here of negative one and positive one. Those are relatively easy values to plug back into that first derivative. So plugging these into our first derivative, we want to find \( f'( \text{-1} )\) and \( f'( \text{1} )\). Now plugging that negative one into this first derivative here, that's going to be \( 3 \times (\text{-1})^2/ (2 \times \sqrt{(\text{-1})^3 + 8}) \).

Now thinking about what's happening with these values on the top, I end up with a positive number. On the bottom, \( \text{-1}^3 + 8 \) is positive. Taking the square root, I still am left

Summary of Curve Sketching Example 3

In this problem, we're asked to sketch the graph of some function with the following properties. Now the properties that we're given here are that our function should be continuous and differentiable everywhere. We're given specific x intercepts, then we're given information about that function's first derivative. Remember, the sign of the first derivative tells us whether a function is increasing or decreasing. We're also given information about that function's second derivative.

Remember, that tells us where a function is concave up or concave down. So let's get started here. Now we're going to keep this first property in mind as we continue on with our other properties. But starting here, we see that our function has x intercepts at x equals 0 and x equals 2. I can go ahead and just plot those points right on my graph here, x equals 0 and x equals 2.

So I have that property all done. Then moving on to my first derivative here, I'm told that this first derivative is greater than 0, so it is positive from negative infinity to 1, and from 2 on to infinity. When that first derivative is positive, that tells me that my function is increasing there. Then where I'm told that it's less than 0 from 1 to 2, that tells me my function is decreasing there. So I don't want to do anything in terms of sketching yet, but I'm going to go ahead and mark on my graph from negative infinity to positive 1.

My function should be increasing. Then from that 1 to 2, my function is decreasing. Then from 2 on to infinity, it begins increasing again, but I don't want to sketch anything until I know what's going on with my concavity. This helps me determine the shape of my graph and how it curves. Now here we're told that that second derivative is greater than 0, it is positive.

When x is less than 0.5 and when x is greater than 1.5. Now when that second derivative is positive, that tells us that our function is concave up. Then when it's negative, less than 0, that tells us our function is concave down. So based on the information we're given here, when x is less than 0.5, so that means from negative infinity until we reach 0.5, we know that our function is concave up. So let's come to our graph here.

From negative infinity to 0.5, I know that my function is concave up. Remember, concave up looks like a smile. Concave down looks like a frown. We know that from 0.5 to 1.5, our function is concave down. So from that 0.5 to 1.5, we know that it is now concave down.

And then finally, from 1.5 on to infinity, we're told that this function is then concave up. So now that we know information about our function's concavity and where it's increasing or decreasing, let's actually start graphing. Now since we know where our function is going to change concavity, it can be helpful to actually just mark where you want your inflection points to be. Now remember, these points can be anywhere. All of the points on your graph can be anywhere as long as your function adheres to the properties that are given.

So we know that our function changes concavity from up to down at x equals 0.5. Now knowing that my function will be increasing as we get to this point and then decreasing until we increase again at this point, I'm going to go ahead and make these points up above the x-axis. So at 0.5, I'm going to go ahead and just mark my point of inflection right here. I know that it changes inflection again from down to up at x equals 1.5, so I'm going to mark another point of inflection at the same y value just because that's where I want it to be. Remember, these points can be anything as long as your final graph adheres to these properties.

Then since I know that my function changes from increasing to decreasing and from decreasing to increasing at these points, that indicates to me local maxima and local minima. So from increasing to decreasing, I know that this is going to be a local maximum. So I'm going to go ahead and mark that higher point right here. I know that it is above my other points because it is a local maximum. Then my function goes from decreasing to increasing at x equals 2.

So I know that this indicates a local minimum. So I'm going to go ahead and mark that point as well, but it is actually already here. It is my x intercept. I know that that means my function will end up looking like some sort of valley like this, whereas my local maximum will be some sort of peak like that. Now that we have some point smarts, it's going to be a little bit easier to sketch our curve.

Now from left to right, I know that my function starts out being concave up and increasing. So starting out being concave up, remember concave up looks like a smile. We are increasing concave up, changing concavity at that inflection point. I'm going to go ahead and try to connect these a little smoother. We're increasing.

We're concave up. We change concavity to being concave down. Remember, concave down looks like a frown, forming this peak at the top, then changing concavity again to being concave up until we finally hit that local minimum and then shoot back up I know that this function is continuous and differentiable everywhere, so I'm going to make those ends go off to infinity. And we want to double-check with all of the criteria that we had here.

Our function is continuous and differentiable everywhere, it checks off that box. I don't have any points where it wouldn't be differentiable and it is definitely continuous. We have x intercepts at x equals 0 and x equals 2. We can see that on our graph there. Then we can see that that first derivative is positive from negative infinity to 1, so it is increasing.

Then it is decreasing from 1 to 2 and increasing again from 2 on to infinity. So it checks off that property as well or both of those properties for the first derivative. Then for the second derivative, the concavity, I can see that my function is concave up looking like a smile. Then it begins to look like a frown from 0.5 to 1.5 and then again is concave up from 1.5 on to infinity. So we can check those last two properties off.

Now if you tried this problem on your own before checking with me, your graph might look a little bit different and that's totally okay. We might not have identical graphs, but we should have these identical properties happening on our graph. Now if you have any questions here, feel free to let us know, and let's keep going.