Finding Global Extrema: Study with Video Lessons, Practice Problems & Examples

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 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) 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). Using the power rule, taking the derivative of 3x2+1 is just going to give me 6x. So 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 any questions, and let's continue practicing in the next video.

Finding Global Extrema (Extreme Value Theorem) Example 1

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 an 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 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. Right? So looking at this function, because it is a basic polynomial function, I know that it is continuous everywhere, and I am given this explicit closed interval with those endpoints included, so I know that it will absolutely have an absolute max and an absolute min. So we can go ahead and find those values using our steps here.

The first step that we want to take is finding our critical points where our derivative is either equal to 0 or does not exist. Now because this is a basic polynomial function, I already know that its derivative will also be a basic polynomial function, so I don't have to worry about anywhere that that derivative does not exist because there is nowhere where that happens. So let's go ahead and find the derivative here and then set it equal to 0. So finding f prime of x here, I'm going to use the power rule. So that first term, that 2 to the power of 4, becomes 8³.

Then that second term minus 8³ is going to become minus 24x². Then that next term gives me a minus 32x. And then that last term is a constant. It goes away. Now looking at this derivative, I see that all of these terms have some common things that I can factor out here.

All of these terms have an x, and they also all have a common factor of 8. So if I pull an 8x out of here, this leaves me with x² minus 3x minus 4. Remember, it's always a good idea to go ahead and factor your derivative once you get it because it will make our lives easier when actually finding those critical points. Now I can do a little bit more factoring here. That x² minus 3x minus 4 is going to factor to (x - 4)(x + 1).

So now that I'm here, I can go ahead and set this equal to 0. Now in doing that, I'm effectively going to be just setting each of these individual factors equal to 0, which is why this made my life easier. I'm going to take 8x, set it equal to 0, x - 4 equals 0 and x + 1 equals 0. Now solving each of these for x, that first one gives me x equals 0. That 8 just goes away.

Then over here, if I add my fours, that gives me x equals 4. And then finally subtracting this one gives me my final critical point, x equals negative one. So I have 3 critical points here, x equals 0, x equals 4, and x equals negative one. So step 1 is done. Now we want to proceed to step 2.

We want to plug both our critical points and our endpoints into our original function f of x. Now remember, we only want to plug our critical points in if they are within that closed interval that has been given to us. Now here, 0, 4, and negative one are all enclosed within that interval from negative 2 to 5, so that means we need to plug all of them back into that original function. So we're going to be working with 5 different values that we have to plug into our function here. All of our critical points, that's x equals 0, x equals 4, and x equals negative one.

And our 2 endpoints, that's x equals negative 2 and x equals positive 5. We want to plug all of these into our function, all of our critical points and our 2 endpoints here. So starting here with x equals 0, plugging that into my original function. Remember, we're working with f of x here, no longer f prime of x. So I'm plugging a 0 in here.

F of 0 is going to be 2 times 0⁴ minus 8 times 0³ minus 16 times 0² plus 3. Now all of those first three terms are just going to cancel out to 0 because 0 is multiplying each of them. So that just leaves me with 3 for my value f of 0. Then for 4, I want to do the same thing here, plugging 4 into my function. 2 times 4⁴ minus 8 times 4³ minus 16 times 4² plus 3.

Now if we multiply all of this out and add this together, this will end up giving me a value of -253 for f of 4. Now from here, we want to do the same thing for all of our remaining values. We're going to plug in negative 1 and negative 2, and you should go ahead and try to do this on your own and then check back in with me. Okay, once you've plugged all of those values into your function, this is what you should have gotten. For f of negative one, your function value is -23.

Then for f of negative 2, we get a value of 35. And finally for f of 5, we get -147. So now with step 2 done we can proceed to step 3 where we're actually going to identify our global max and global min. Remember that your largest value is going to be your global maximum whereas your smallest value is your global minimum. So let's go ahead and identify our largest and smallest values out of these 5 function values that we just found.

Now here, I can see that my largest value out of all of these, making sure that I'm paying attention to the sign here, is this positive 35. So this tells me that my function reaches a global maximum of 35 at x equals negative 2. So that is my global maximum. Then for our global minimum, we want to look for our very smallest value. And I see that looking at all of these, my absolute smallest value is this negative 253.

So this tells me that my function then reaches a global minimum of -253 at x equals positive 4, and that's all there is to it. Let us know if you have questions, 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 2

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 maximum or minimum 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 maximum or minimum 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 maximum or minimum. 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 maximum and minimum, 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 going to 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 onto 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, a y prime, that's going to be equal to 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 going to 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 a 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 going to 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 3

In this problem, we're asked to find the absolute maximum and minimum values of the function over the given interval. The function we are working with is \( f(x) = x \cdot \log(x) \) over the interval from 1 to 2, with both of those endpoints included. This is a closed interval, indicated by the square brackets. Based on the extreme value theorem, this function does have an absolute maximum and minimum, which we can find by examining the function carefully. The first step is to find the critical points.

This involves where the derivative is either equal to 0 or does not exist. First, we need to find the derivative \( f'(x) \). Since \( f(x) \) involves one function \( x \) multiplying another function, the natural log of \( x \), I will use the product rule. The derivative of \( \log(x) \) is \( \frac{1}{x} \).

Using the product rule gives:

\( f'(x) = x \cdot \left(\frac{1}{x}\right) + \log(x) \cdot 1 \)

After simplifying, we have:

\( f'(x) = 1 + \log(x) \)

We now need to look for where this derivative equals 0 or where it does not exist. The derivative does not exist where \( x = 0 \), but since 0 is not in the interval, it is irrelevant here. Setting the derivative equal to 0, we find:

\( \log(x) = -1 \)

Raising \( e \) to the power of both sides gives:

\( x = e^{-1} \) or \( x = \frac{1}{e} \).

This value \( \frac{1}{e} \approx 0.37 \) is outside the interval from 1 to 2, so it is also not relevant for our analysis.

Since none of the critical points are within the interval, we only need to consider the endpoints. Plugging in these endpoints into the original function \( f(x) = x \cdot \log(x) \) gives:

\( f(1) = 1 \cdot \log(1) = 0 \)

\( f(2) = 2 \cdot \log(2) \approx 1.386 \)

Only these two values need to be considered. The largest is 1.386 at \( x = 2 \), representing the global maximum over the interval. The smallest, 0 at \( x = 1 \), represents the global minimum.

Therefore, we have our global minimum of 0 at \( x = 1 \) and our global maximum of approximately 1.39 at \( x = 2 \). If you have any questions, please let us know, and I'll see you in the next video.