In my last lesson, I introduced optimization problems and I discussed local extrema. You should check that out if you haven’t already. The next thing that I would like to discuss now is finding global maximums and minimums. The first step in finding a global maximum or minimum of a function is actually very similar to finding the local max and min values.

#### But once you know about the local maximums and minimums, how do you find the global extrema?

Finding the global extrema from the local extrema is really quite simple. And there really is only one way to find the global maximum and minimum values of a function. You just need to find a list of all possible *x *values where the global max or global min may occur. Then once you have created a list of all possibilities, you just plug them all into the original function.

Not the function’s derivative or the function’s second derivative. **But the original function.**

You will test for global extrema of* f(x)* using *f(x)*, not *f'(x)* or *f”(x)*. This is after you have your list of all possible locations where the global extrema could occur (which will require the use of *f'(x)*). But my point is that there wouldn’t be a first derivative test or a second derivative test with the global extrema like there was with finding local extrema.

## So what does this look like in practice?

Let’s use an example. For example, let’s say that we are asked to find the global maximum and the global minimum of on the domain .

Notice we are being asked to find the global extrema on a specific domain. This is important because a lot of functions either go to infinity or to negative infinity as *x* either gets infinitely large, infinitely small, or approaches some specific value. So if we were asked to find the global maximum of a function that goes to infinity as *x* goes to infinity, we wouldn’t be able to do this. There would be no maximum since the function only continues to grow.

So we know we will be limited to a specific domain.

As I said before, finding global extrema starts out exactly like finding local extrema. The first thing we need to do is find the critical values of our function. To do this, we just need to find its derivative and set and solve for *x*.

$$f(x)=2x^3-\frac{5}{2}x^2+x-1$$ $$f'(x)=6x^2-5x+1$$ Then set $$6x^2-5x+1=0$$ and solve for *x* to find the critical values. To do this, we can factor the left side of the equation. $$(3x-1)(2x-1)=0$$ To solve this we can set each factor equal to zero individually. $$3x-1=0 \ \ \ \ and \ \ \ \ 2x-1=0$$ $$3x=1 \ \ \ \ and \ \ \ \ 2x=1$$ $$x=\frac{1}{3}, \ \frac{1}{2}$$

So now we know that this function has two critical values, and . Now this is where things get different with a global max/min problem versus finding the local max/min. **We also need to consider the endpoints of our given domain as critical values!**

This will always be the case when we are looking for a global maximum or minimum. The problem asked us to find the global extrema on the domain . Therefore, we will also say that and will be treated as critical values that we need to test.

### So how to we test our critical values?

The first thing that I would like to do is list out all of the *x *values we will be testing in one place. Remember, the list of values we need to test came from two places:

- Setting and solving for
*x*. - Each of the endpoints of the domain on which we need to find the global maximums and minimums.

So in this case we’ll have four total *x *values that we need to test: $$x=-4, \ \frac{1}{3}, \ \frac{1}{2}, \ and \ 5$$ To test these points, all we need to do is plug each of the four points into *f(x)*. Whichever on outputs the largest number will tell us the global maximum. Whichever outputs the smallest number will tell us the global minimum.

$$f(-4)= \ 2(-4)^3-\frac{5}{2}(-4)^2+(-4)-1 \ = -173$$ $$f \bigg( \frac{1}{3} \bigg) = \ 2 \bigg( \frac{1}{3} \bigg) ^3-\frac{5}{2} \bigg( \frac{1}{3} \bigg) ^2+ \bigg( \frac{1}{3} \bigg) -1 \ = -\frac{47}{54}$$ $$f \bigg( \frac{1}{2} \bigg) = \ 2 \bigg( \frac{1}{2} \bigg) ^3-\frac{5}{2} \bigg( \frac{1}{2} \bigg) ^2+ \bigg( \frac{1}{2} \bigg) -1 \ = -\frac{7}{8}$$ $$f(5)= \ 2(5)^3-\frac{5}{2}(5)^2+(5)-1 \ = \frac{383}{2}$$

So we can see that the smallest of these four numbers is *-173* and the largest of them is . Therefore, the global maximum of *f(x)* on is which occurs when . And the global minimum of* f(x)* on is *-173* which occurs when .

We can even graph our function using Desmos along with the critical points to make sure our answer makes sense. You can click on the link in the last sentence to see a larger version of the graph.

## Extra practice

If you’d like some extra practice finding global maximums and minimums, here are a few examples you can work through on your own. A couple of these examples will require the use of the product rule and quotient rule.

For each of the following, find the global maximum and minimum of the given function on the given domain or explain why one doesn’t exist.

$$f(x)= 2x^4 + 5x^2 – 12x \ \ \textrm{on the domain} -1 \leq x \leq 2$$ $$g(x)= xe^x +6x^3-12 \ \ \textrm{on the domain} -3 \leq x \leq 0$$ $$h(x)= \frac{x^4-3x^2+1}{x+1} \ \ \textrm{on the domain} -2 \leq x \leq \frac{3}{2} $$ $$j(x)= \frac{x^4-3x^2+1}{x+1} \ \ \textrm{on the domain} -\frac{9}{10} \leq x \leq \frac{3}{2}$$

Hopefully all of this helps with global maximums and minimums, but as always I’d love to hear your questions if you have any. If you find that you get stuck as you’re working through some of these extra practice problems just let me know. Just email me at **jakesmathlessons@gmail.com** and I’ll see if I can help provide a bit more clarification. **You can also use the form below to subscribe to my email list and I will send you my bonus FREE calculus 1 study guide to help you survive calculus!** Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about derivatives as well.