- In this section, we see how to employ differentiation to help find minimum and maximum values.
*Definition.*Let \(f\) be a function with domain \(D\), and let \(c\) be a point in \(D\).- Then \(f(c)\) is the
*absolute minimum value*of \(f\) on \(D\), if \(f(c)\le f(x)\) for all \(x\) in \(D\). - \(f(c)\) is the
*absolute maximum value*of \(f\) on \(D\), if \(f(c)\ge f(x)\) for all \(x\) in \(D\). - \(f(c)\) is a
*local minimum value*, if \(f(c)\le f(x)\), when \(x\) is near \(c\). - \(f(c)\) is a
*local maximum value*, if \(f(c)\ge f(x)\), when \(x\) is near \(c\). - "\(x\) is near \(c\)" means: the property holds for \(x\) in some open interval containing \(c\).
- The maximum and minimum values of \(f\) are also called
*extreme values*. - The absolute extreme values of \(f\) are also called
*global*extreme values.

- Examples.
- The function \(f(x)=x^2\) has \(f(0)=0\) as a global minimum value, but it does not even have a local maximum value.
- The function \(f(x)=sin(x)\) has \(f(\frac{\pi}{2})=1\) as a global maximum value, and \(f(-\frac{\pi}{2})=-1\) as a global minimum value. Note that the same global extreme value can be taken up more than once. Here, we have \(f(\frac{\pi}{2}+2k\pi)=1\) for any integer \(k\).
- We'll see soon how to verify that the function \(f(x)=(x+1)x(x-1)=x^3-x\) has \(f(-3^{-1/2})=-3^{-3/2}+3^{-1/2}\) as a local maximum value, and \(f(3^{-1/2})=3^{-3/2}-3^{-1/2}\) as a local minimum value. It does not have no global extreme values.

*The extreme value theorem.*- Let \(f\) be a function, which is continuous on its domain a closed interval \(D=[a,b]\). Then \(f\) attains an absolute maximum value \(f(c)\), and an absolute minimum value \(f(d)\) at some numbers \(c,d\) in \(D\).
*Sketch of proof.*Let \(\mathrm{Im}(f)\) be the collection of the values \(f(a)\) for \(a\) in \(D\). It can be pictured as the projection of the graph of \(f\) to the \(y\)-axis. It is called the*image*of \(f\).- Since \(D\) is a closed interval, and \(f\) is continuous, its image \(\mathrm{Im}(f)\) is also a closed interval \([\alpha,\beta]\).
- Then by construction \(\alpha=f(c)\) is a global minimum value, and \(\beta=f(d)\) is a global maximum value. QED (End of proof)
- This is only a sketch, because it's not easy to show that the image \(\mathrm{Im}(f)\) is a closed interval.
- Questions like that are answered in the field of Mathematics called
*Topology*. It's a very interesting topic, I suggest learning about it if you have the chance!

*Fermat's theorem.*- Let \(f\) be a function. Suppose that \(f(c)\) is a local extreme value of \(f\), and that \(f'(c)\) exists. Then we have \(f'(c)=0\).
- Here the proof is easier, you can check it out in the textbook if you're interested.

- Examples.
- Let \(f(x)=x^2\). Then \(f'(x=0)=(2x)(x=0)=0\), and \(f(x=0)\) is a global minimum value.
- Let \(f(x)=x^3-x\). Then \(f'(x)=3x^2-1\), the zeroes of which are \(x=\pm3^{-1/2}\). These are local extreme values.
- Let \(f(x)=x^3\). Then \(f'(x=0)=(3x^2)(x=0)=0\), but \(f(x=0)\) is not a local extreme value.
- Let \(f(x)=|x|\). Then \(|0|\) is a global minimum value, but \(f'(x=0)\) does not exist.

- Summary of examples
- It is possible that \(f'(c)=0\) even though \(f(c)\) is not a local extreme value
- It is also possible that \(f(c)\) is a local extreme value, even though \(f'(c)\) does not exist.

*Critical numbers**Definition.*Let \(f\) be a function with domain \(D\), and let \(c\) be a number in \(D\). Then \(c\) is a*critical number*of \(f\), if \(f'(c)=0\), or \(f'(c)\) does not exist.- Using this definition, Fermat's theorem reads as follows.
- If \(f\) has a local extreme value at \(c\), then \(c\) is a critical number of \(f\).

- Let \(f\) be a function with domain a closed interval: \(D=[a,b]\).
- The extreme value theorem is telling us that since \(f\) is defined on a closed interval \([a,b]\), it has a global minimum and a global maximum.
- Even though Fermat's theorem doesn't find all the local extreme values at once, it decreases their number: for \(f(c)\) to be a local extreme value, we need \(c\) to be a critical number of \(f\).
- It can also happen that a global extreme value is taken up at one of the endpoints of the closed interval.

*The closed interval method.*- To find the global extreme values of \(f\), we can proceed as follows.
*Step 1.*Find the values of \(f\) at its critical numbers.*Step 2.*Find the values of \(f\) at the endpoints of its domain \([a,b]\).*Step 3.*The global minimum and maximum values of \(f\) are going to be among the values found in Steps 1 and 2.

- Exercises. 4.1: 48, 52, 54, 56, 58, 60, 62, 70