Finding Global Extrema - Online Tutor, Practice Problems & Exam Prep

Critical Points

At this point in the course, we know how to find the local extrema of a function by looking at its graph. For example, for a function g(x), we can see that it has a local maximum at x equals negative 4, a local minimum at x equals negative 1, and finally, a local maximum when x is equal to positive 4. But we won't always be given the graph of a function, and we'll still need to identify the local extrema when just given the function itself. But how exactly can we even start doing that? Well, the first step in being able to find the local extrema of a function is being able to find what's called that function's critical points.

Now this is exactly what I'm going to walk you through here, what critical points are and how exactly to find them. So let's go ahead and jump right in. Now coming back to our graph of g(x), if we look at all of these local extrema, they all have something in common if we think about their tangent lines. For this first local maximum and minimum, if I draw a tangent line, I can see that it is a horizontal tangent line. And for this last local maximum over here, because it comes up to a point, I know that it has no tangent line at all and this is actually going to be true for all local extrema.

They either have a horizontal tangent like those first two points or no tangent line at all. So why does this information matter? Well, if we know information about a function's tangent, then we also know information about that function's derivative because it's just the slope of the tangent. So if a function has a horizontal tangent line, I know that at that point, my derivative will be equal to 0, that zero slope of that horizontal tangent. And if my function has no tangent line at a point, I know that there, my derivative does not exist.

Now these points where a derivative equals 0 and where a derivative does not exist are what's called critical points. And remember earlier, I mentioned that finding critical points is going to be the first step in being able to find local extrema. So let's actually find the critical points of this function here, f(x). Here we have x³ - 12x + 5. And in order to find those critical points, that means we need to find where our derivative equals 0 and where our derivative does not exist.

So in order to do that, let's first just find what our derivative is. So f'(x) here using the power rule is going to be 3x² - 12. So now that I have that derivative, to find where it's equal to 0, I just need to set that 3x² - 12 equal to 0 and solve for x. Now adding 12 on both sides here, that will cancel out, leaving me with 3x² = 12. Then dividing both sides by 3 cancels that out.

I'm left with x² = 4. And then finally, the last step, taking the square root on both sides, I end up with x is equal to plus or minus 2. So this gives me two critical points where x is equal to positive 2 and where x is equal to negative 2. But remember, this is just half of the story. We also need to find where a derivative does not exist.

And this is specifically where it does not exist for values of x that actually exist in the domain of our original function. Now here, since we're working with a basic polynomial function, there is actually nowhere where this derivative does not exist. But it is something to keep in mind as you work with different types of functions like, say, rational functions or those with rational exponents. So here we already have all of our critical points. We have two critical points where x is equal to positive 2 and where x is equal to negative 2.

So now that we have our critical points, there is something that I want to note here, and that's that not all critical points are local extrema. So what exactly does that mean? Well, it means that just because positive two and negative two are critical points, that doesn't automatically make them a local maximum or local minimum. Now, this can be a bit hard to understand, so let's come back to our graph here. We know that these local extrema have either a horizontal tangent or no tangent at all.

But this point here at x equals 2 also has a horizontal tangent, and it's not a local max or min at all. So just because a function has a critical point, meaning that its derivative is equal to 0 or does not exist, that doesn't automatically make it a local maximum or local minimum. Finding critical points is just the first step in finding local extrema and also in finding a lot of other things out about a function. Critical points are actually going to be used to find out a lot of different things, including local and global extrema, but also a lot of other stuff. Now I'm going to walk you through absolutely everything that critical points are used for as we continue in this course.

But for now, let's get some more practice with finding critical points. I'll see you in the next video.

Critical Points Example 1

In this problem, we're asked to find the critical points of the given function \( f(x) \). The function that we're given here is \( f(x) = \frac{x^2}{x - 4} \). Remember that when we're asked to find critical points, that means we want to find where our derivative is equal to 0 or where it does not exist. So, we need to start by first finding the derivative of our function. Now, because we're working with a rational function, we have one function being divided by another; we need to use the quotient rule to find our derivative \( f'(x) \), which can be remembered using the memory tool for the quotient rule: "low d high minus high d low over the square of what's below."

Let's derive it: \[ f'(x) = \frac{(x-4) \cdot 2x - x^2 \cdot 1}{(x-4)^2} = \frac{2x^2 - 8x - x^2}{(x-4)^2} = \frac{x^2 - 8x}{(x-4)^2} \] We can factor out an \( x \) in the numerator, giving us: \[ f'(x) = \frac{x(x-8)}{(x-4)^2} \] It's a good idea to factor if you see something that's easily factorable because it will help you in solving the problem.

We're not done yet; this is just our derivative. We want to find where this derivative is equal to 0 or where it does not exist. To find where the derivative is equal to 0: Setting the numerator equal to 0, we have \( x(x-8) = 0 \). Solving this gives \( x = 0 \) and \( x = 8 \). So, we have two critical points: \( x = 0 \) and \( x = 8 \).

Now, let's consider where the derivative does not exist. Since this is a rational function, we need to recognize that the derivative will not exist where the denominator equals zero. If we set \( x-4 = 0 \), we find \( x = 4 \). However, the original function is also undefined at \( x = 4 \) (as \( x - 4 \) is in the denominator of \( f(x) \)), thus \( x = 4 \) is not a critical point because the function itself does not exist there.

Therefore, for the function \( \frac{x^2}{x-4} \), there are two critical points: \( x = 0 \) and \( x = 8 \). Feel free to let us know if you have any questions, and let's continue practicing.

Find the critical points of the given function.

$g(x)=\frac13x^3-\frac12x^2-12x$

• A. $x=0,x=12$
• B. $x=-4,x=3$
• C. $x=-6,x=2$
• D. $x=4,x=-3$

Find the critical points of the given function.

$f(t)=\frac{6t}{t^2+1}$

• A. $t=0,t=6$
• B. $t=1$
• C. $t=-1,t=1$
• D. No critical points

Find the critical points of the given function.

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

• A. $x=0$
• B. $x=\sqrt2,x=-\sqrt2$
• C. $x=0,x=\sqrt2$
• D. No critical points

Critical Points Example 2

Here, we're asked to find the critical points of the given function, and we're also given a specified domain for this function. So we're looking at the function \( f(\theta) \), which is equal to \( 2 \sin(\theta) + \cos^2(\theta) \), and we're looking at this over the interval from \( 0 \) to \( 2\pi \). So let's go ahead and get started here. Remember that in order to find the critical points, we want to find where the derivative, in this case, \( f'(\theta) \), is either equal to 0 or it does not exist. So let's go ahead and start by finding our derivative.

Finding my derivative here, \( f'(\theta) \), for that first term \( 2 \sin(\theta) \), the derivative of \( 2 \sin(\theta) \) is going to be \( 2 \cos(\theta) \). Then taking the derivative of that second term, it can be helpful to think of this as the cosine of theta squared so that it's more clear how we're using the chain rule here. This is going to be equal to \( 2 \cos(\theta) \times (-\sin(\theta)) \). Let's do a bit of simplification here.

This is \( 2 \cos(\theta) - 2 \cos(\theta) \sin(\theta) \). Now remember that it's always going to be useful to factor where you can. And here, since both of these terms have a \( 2 \cos(\theta) \), I'm going to go ahead and factor that out. So that gives me \( 2 \cos(\theta)(1 - \sin(\theta)) \). Now remember that in order to find our critical points, we want to now set this equal to 0.

Setting this equal to 0, I am effectively setting each of these individual terms equal to 0. So that means I'm looking for where \( 2 \cos(\theta) \) is equal to 0 and also where \( 1 - \sin(\theta) \) is equal to 0. Starting with this over here, I can then just cancel that 2. So I'm effectively just looking for where the \(\cos(\theta)\) is equal to 0. And thinking about my unit circle, this is where any of our x-values are going to be 0.

Remember, we're specifically looking over the domain from \(0\) to \(2\pi\). So we don't need to consider every single variation of that because we know that it will continue being 0 every time we go around the unit circle. We only want to have those points that are in that domain from \(0\) to \(2\pi\). So that is going to be where \(\theta\) is equal to \(\pi/2\) or \(3\pi/2\). Now, if you need a refresher on anything regarding trig functions, remember that you can look back at some older content.

But just as a reminder, remember that when we're looking at something like the cosine of \(\theta\) being equal to 0, we want to think, okay, what value of \(\theta\) makes the cosine 0? So those values in this case are \(\pi/2\) and \(3\pi/2\). So those are 2 critical points. Now let's move on to the other side here. We have 1 minus \(\sin(\theta)\).

Now if I add the \(\sin(\theta)\) to both sides, I am then looking for where the \(\sin(\theta)\) is equal to 1, just swapping those sides to make this a bit more clear. So remember, in this case, we're looking for, okay, what value does \(\theta\) need to be in order for the sine to be 1? Thinking back to our unit circle, okay, you can also think of sine as being your y-value if you're visualizing that coordinate system. So in this case, remember, we are still looking only from \(0\) to \(2\pi\). This is going to be where \(\theta\) is equal to \(\pi/2\).

And that gives us our other critical point. It overlaps with what we already found over here. So effectively, we only have 2 critical points where \(\theta\) is equal to \(\pi/2\) and where \(\theta\) is equal to \(3\pi/2\). Now remember that our critical points are also where our derivative does not exist, but there are no values of \(\theta\) within our specified domain for which that would happen, so we don't have to worry about that. Thanks for watching and I'll see you in the next one.

Finding Global Extrema (Extreme Value Theorem)

Based on everything that we've learned so far, we know how to identify the global maximum and minimum values of a function by looking at its graph. We know that these are the absolute highest and lowest points that we see our function reach. But we won't always have the graph of a function to rely on and we'll still need to be able to find the global extrema just based on the function itself. And that's exactly what I'm going to show you how to do here, how to find the global maximum and minimum of a function following a step by step process, starting first with the extreme value theorem. So let's go ahead and jump in here.

Now, before we can actually find the global extrema of a function, we first have to determine if that function even has a global max and global min. And that's what we can do using the extreme value theorem because we know that if we had a parabola that just went off to infinity, that function doesn't actually have a global maximum. So the extreme value theorem tells us that if a function f is continuous, meaning it doesn't have any breaks or holes in it just like we see in this function here, just a smooth continuous curve, and it is continuous on a closed interval from a to b, that means both a and b are included within that interval. Then if both of those things are true, then we know that our function f has a global maximum and a global minimum value within that interval from a to b. So really, we're checking for 2 things.

Is our function continuous, and is it defined over a closed interval? If those two things are true, we know that our function has a global maximum and global minimum value. So with our function f ( x ) = 3x2 + 1, if we want to find if this function has a global maximum and minimum, we want to check those 2 things. Is this function continuous? Well, this is a basic polynomial, so I know that it is definitely continuous.

And is it defined over a closed interval? When we're asked specifically to look at the interval from negative 2 to 4, we can see that this interval is closed based on those square brackets, so this is also true. So, yes, this function f ( x ) does have a global maximum and global minimum value. Okay. Now that we know it has those values, how can we actually find them?

Once you verify that those values exist, you can then actually find your global maximum and minimum by testing the critical points and the endpoints of your function. These will then give you your global max and your global min. So let's actually work through this process step by step with our function f ( x ) = 3x2 + 1 here. The very first thing that we want to do is find the critical points of our function. Now, in order to do that, we need to find that derivative and set it equal to 0 and also determine where it does not exist.

Now since we're working with this polynomial here, I want to find f'(x) = 6x. To find my critical points here, I just need to set this equal to 0 because there is nowhere that this derivative does not exist. Now setting this equal to 0 gives me my single critical point x = 0. So with step 1 done, we can move on to step 2 and now plug our critical points and our endpoints into our original function f ( x ).

One thing to note here is that you only want to plug critical points in that are inside of your closed interval. Now here, our closed interval is from negative 2 to 4, and 0 is within that interval. So I do want to proceed in plugging that in. So I want to plug 3 values in here, x = 0, x = -2, and x = 4. That's my critical point and my 2 endpoints.

I'm plugging these back into my original function f ( x ). So plugging 0 in here to that 3x2 + 1, that's going to give me 3 × 02 + 1, which is just equal to 1. Then plugging that negative 2 in, that's 3 × ( -2 )2 + 1, which gives me a value of 13. And then finally, plugging 4 in, that's 3 × 42 + 1, which gives me a final answer here of 49. So I've plugged all of these values in.

We have completed step 2, and we can proceed to our final step where we're actually going to get our global max and global min. Now of the values that we just found in step 2, we want to look for our largest and our smallest values. Our largest value is going to give us our global maximum. Now looking at these three values, I have 1, 13, and 49. This 49 is clearly the largest value, meaning that my function has a global maximum of 49 at x = 4.

Then, of course, our smallest value will then be our global minimum. Here, my smallest value I can see is 1, so that means that my function then has a global minimum of 1 at x = 0. So, in order to find the global extreme of a function, you first have to verify if you even have a global max and global min using the extreme value theorem. From there, you want to find your critical points. Then plug your critical points and endpoints back into your original function so that you can identify your global max and global min.

Feel free to let us know if you have questions, and let's continue practicing in the next video.

Finding Global Extrema (Extreme Value Theorem) Example 3

In this problem, we're asked to find the absolute maximum and absolute minimum values of the function over the given interval. Now, the function that we're working with here is f(x)=2x4-8x3-16x2+3. And the given interval is a closed interval from negative 2 to positive 5. Now, in these problems, we're asked to find the absolute or global maximum and minimum values. Remember, we first want to think about the extreme value theorem, which tells us whether a function actually has an absolute max and absolute min over a closed interval.

Now, this is really more of a formality at this point because in a vast majority of these problems where you're given a function and an explicit closed interval to work with, there will absolutely be an absolute max and an absolute min. But we still want to make sure that we're at least thinking about the extreme value theorem. Thus, looking at this function, because it is a basic polynomial function, I know that it is continuous everywhere, and I have been given this explicit closed interval with those endpoints included, so I know that it will absolutely have an absolute max and an absolute min. We can go ahead and find those values using our steps here.

Now, the first step that we want to take is finding our critical points where the derivative is either equal to 0 or does not exist. Because this is a basic polynomial function, I already know that its derivative will also be a basic polynomial, so I do not have to worry about anywhere that this derivative does not exist because there is nowhere where that happens. Let's go ahead and find the derivative here and then set it equal to 0. Finding f'(x), I am going to use the power rule. That first term becomes 8x3.

Then that second term becomes -24x2. The next term yields -32x. The last term, being constant, disappears. Reviewing this derivative, I note that there are common factors I can pull out here.

All of these terms share a common x, and they all possess a common factor of 8. If I extract an 8x out of this scenario, it leaves me with x²-3x-4. Factoring further, x²-3x-4 factors into (x-4)(x+1).

My next action is to set this equal to 0, which effectively means setting each of these individual factors to 0. Solving this, I find that x=0, x=4, and x=-1. These are our critical points.

Next, we plug both our critical points and our endpoints into the original function f(x). We ensure to include only those critical points within the interval [−2, 5]. Thus, plug in x=-2, x=-1, x=0, x=4, and x=5 into our function. After evaluating, our values for the function are as follows:

• f(-2)=35
• f(-1)=-23
• f(0)=3
• f(4)=-253
• f(5)=-147

Identifying the global maximum and minimum, we find the largest value is 35 at x=-2, making this our global maximum. The smallest value is -253 at x=4, setting this as our global minimum.

If you have questions, let us know, and let's keep practicing!

Find the global maximum and minimum values of the function on the given interval. State as ordered pairs.

$y=x+\frac{2}{x};[0.25,3]$

• A. Global maximum at $\left(0.25,8.25\right)$, Global minimum at $\left(0,2\right)$
• B. Global maximum at $\left(0.25,8.25\right)$, Global minimum at $\left(\sqrt2,2\sqrt2\right)$
• C. Global maximum at $\left(0.25,8.25\right)$, Global minimum at $\left(3,3.67\right)$
• D. Global maximum at $\left(\sqrt2,2\sqrt2\right)$, Global minimum at $\left(-\sqrt2,-2\sqrt2\right)$

Find the global maximum and minimum values of the function on the given interval. State as ordered pairs.

$y=8+27x-x^3;[0,4]$

• A. Global maximum at $\left(3,62\right)$, Global minimum at $\left(0,8\right)$
• B. Global maximum at $\left(3,62\right)$, Global minimum at $\left(4,52\right)$
• C. Global maximum at $\left(-3,-46\right)$, Global minimum at $\left(0,8\right)$
• D. Global maximum at $\left(4,52\right)$, Global minimum at $\left(-3,-46\right)$

Finding Global Extrema (Extreme Value Theorem) Example 3

In this problem, we are asked to find the global maximum and minimum values of the function over the interval from negative infinity to positive infinity or state that there isn't a global max or min at all. Now the function that we're working with here is y=x2+6x-3. And you may be wondering here, well, if we don't have a closed interval, how can we know that this function has a global max or a global min at all? And you would be totally right in thinking that because we don't know just looking at this function and based on that interval if it does have a global max or a global min. But here, we can still follow this same process and pair that with our knowledge of functions in order to figure out if there's a global max and a global min, and if so, what those values actually are.

So let's go ahead and dive into this here. Now looking at this function, y=x2+6x+3. That is just a quadratic equation.

So that means that I'm gonna be working with a parabola. And further, because this is a positive x², I know that I'm going to be working with an upward facing parabola, so something that looks like this. Now when we're working with parabolas, we know that both sides go up to infinity. So that means that we're not going to have a global maximum value. But because we have this bottom point, this little valley that happens here, we are going to have a global minimum.

We have a lowest point on our function even though it's over this open interval. And we can follow these same steps in order to find what that global minimum value is. So let's get started here with step 1 and find our critical points. This is where our derivative is either equal to 0 or does not exist. So finding our derivative here, y'=2x+6 using the power rule.

And here, since we're working with a basic polynomial or here a linear function, we don't have to worry about where this doesn't exist. It exists everywhere. So we only need to set this equal to 0 to find our critical points. Now subtracting 6 on both sides gives me 2x=-6. Then dividing both sides by 2 gives me x=-3.

So this is my one critical point, x=-3, and we can move on to step 2 here. Now in step 2, we are told to plug our critical points and our endpoints into our original function. Now this might seem kind of strange because we do not have endpoints here, and we also do not have an interval to check that our critical points are inside. But that's okay. Because we already know what's going on with our function and we have our one critical point, we can just plug this critical point into our original function and see what happens.

Now if I plug x=-3 into my original function, that's gonna be y=-32+6×-3-3. If I multiply all this out, add stuff together, that's going to give me a value of negative 12. Now because I only have one point that I'm working with here, you may be wondering, okay. Well, I don't have a largest and I don't have a smallest because I'm only looking at one singular value. But remember, we already know what's going on with our function.

We know that it's a polynomial function with this global minimum point. And this point is where my critical point is because I only have one critical point where that tangent line is horizontal, where that derivative is equal to 0, and I found it. It's x=-3. So when I plug that x=-3 in, this is going to give me my global minimum. So I can already say that this function has a global minimum of negative 12 at x=-3.

And this is my answer here. We do not have a global maximum here, so we can say that there is no global max because it goes off to infinity. We are not working with that closed interval, but we know that we have that global minimum. Now I know that this can be a little bit confusing. Feel free to let us know if you have any questions here, and we're gonna keep practicing coming up.

Now when we're working with an open interval, just know that this is where we need to use our knowledge of functions. Like here, we knew that this was a parabola that was facing upward, so it had to have a global minimum. Thanks for watching, and I'll see you in the next one.

Finding Global Extrema (Extreme Value Theorem) Example 4

In this problem, we're asked to find the absolute maximum and the minimum values of the function on the given interval. Now, the function we're working with here is f(x)=x+cos(x) over the given interval from 0 to 2π (2 pi). This is a closed interval with both endpoints included. So looking at this function and this given interval, I know that this function does have a global max and a global min based on the extreme value theorem. It is continuous over this closed interval.

So let's go ahead and get started with step 1 and find our critical points. This is where our derivative is either equal to 0 or does not exist. So we need to start here by finding our derivative f′(x). Now the derivative of x is just 1, and the derivative of cosine is negative sine, so this derivative is 1-sin(x). So looking at this derivative over this interval, there is not going to be anywhere where the derivative doesn't exist.

Remember, I only care about what's going on inside of the interval, so I just need to find where this is equal to 0. So if I go ahead and add sine on both sides, this will end up giving me that sin(x) is equal to 1. So I need to find any values for x that are in this interval that are equal to 1. Now there are an infinite number of values for which the sine is equal to 1, but since we're only working with that interval, we know that that's just where x is equal to π/2 (pi over 2). So we have our one critical point here, π/2.

Now we can proceed to step 2 and plug our critical points and our endpoints into our original function. So we wanna plug in x=π2 (x equals pi over 2). That's our critical point. And then our 2 endpoints, that's x=0 and x=2π (x equals 2 pi). So we're plugging these into our original function f(x).

So for π/2, that's f(π2). It's going to be π/2 plus the cosine of π/2. Now that gives me here, so π/2 plus cosine of π/2 is going to be π/2 because the cosine of π/2 is just 0. And then for 0, plugging this in, f(0) is equal to 0 +cosine of 0. Now the cosine of 0 is 1 and a 0 is just 0, so this is equal to 1. Then finally, our last value here, our other endpoint, 2 pi, plugging 2 pi in, so that's 2 pi plus the cosine of 2 pi.

The cosine of 2 pi is again 1. So this is 2 pi plus 1. Now for these function values that have pi in them, it's gonna be helpful here to think about them in terms of decimal just so that it's very clear which one is the largest and which one is the smallest. So just know that π/2 is about 1.57 as a decimal and 2π+1 is about equal to 7.28 as a decimal. So now that we have all of these function values, we wanna look for the largest and the smallest.

These give us our global max and global min. Now of these values, I can see that the largest value is this 2π plus 1 or this 7.28. So this is my global maximum right here, global max, global maximum of 2π+one@xequals2π. Then my smallest value here is actually just a one when x=0, so this is my global minimum. So my function has a global minimum of 1@xequals0, and those are absolute maximum and minimum values.

Let's keep going.