# Definitions of absolute and local minima/maxima

• 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

• 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 and critical numbers

• 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 $$f$$.

# The closed interval method

• 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