Lesson 3.2: Minima and Maxima

Graph of y=f(x)=x^5/125-4x^3/15+64x/25
Graph of y=f(x)=x^5/125-4x^3/15+64x/25

The concept of minimum and maximum points is a pretty intuitive one (and one you should remember from Algebra II), and it easily follows that there are “local” minima and maxima—points where the function is at its greatest or smallest value within a certain range of the function’s input. For example, the graph of $y=f(x)$ to the right represents the function $f(x)=\frac{x^5}{125}-\frac{4x^3}{15}+\frac{64x}{25}$ (yes, I know it’s a touch ugly, but there’s a method to my madness; bear with me) and we can see examples of local minima at $x=-2$ and $x=4$, while there are local maxima at $x=-4$ and $x=2$. (Obviously, we can find the actual minimum and maximum values, of course, by evaluating $f(-2)$, $f(4)$, and so forth.)

But the fact that these points are just locally minimum and maximum points presents a slight problem in defining those terms (collectively referred to as “local extrema”). Does it surprise you that we can use calculus not just to find things but also to define them? Well, it surprised me. Sure enough, think about these points in terms of the derivative: the derivative just to the left of $-2$ is negative, and just to the right it’s positive. It seems reasonable to think that it must pass through zero as it changes from negative to positive, doesn’’t it? It does, so we’ll break for a quick aside about…

The Intermediate Value Theorem

The IVT states, basically, that if we have a function $g(x)$ continuous on the interval $x\in [a,b]$, then $g(x)$ takes on every value $v$ such that $g(a) < v < g(b)$. It makes a lot more sense to think about than to write out, but basically it’s saying that continuous functions don’t jump around (duh!). The proof is pretty dumb, but if you’re interested you can read about it on Wikipedia.

Animation showing the progress of a tangent line as it has approaches a zero slope

Getting back to our issue of local extrema, indeed $f'(x)$ does pass through zero. The animation (106 KB; my apologies, as it may take some time to load) to the right shows the progress of the tangent line from $-3$ to $-1$, and you can see that (as well as understand why, I hope) the point where $f'(x)=0$ is at $x=-2$, the minimum point. It makes sense when you think about it, doesn’t it? The situation is identical for maxima, except that the derivative is changing from positive to negative.

So going back to our $f(x)$, now we see why I picked such an ugly function: its derivative is easy: $f'(x)=\frac{x^4}{25}-\frac{4x^2}{5}+\frac{64}{25}$ You don’t see it yet? Oh well, you’re forgiven. My point is we can factor that: $f'(x)=\frac{1}{25}(x+4)(x-4)(x+2)(x-2)$ which makes it really easy to solve $f'(x)=0$ thanks to our nearly-forgotten buddy, the Zero Product Property, which tells us that $f(x)$ has local extrema at $x=\{-4, -2, 2, 4\}$...or does it?

Graph of y=h(x)=x^3
Graph of y=h(x)=x^3

Not definitively. Consider $h(x)=x^3$ at $x=0$: $h'(x)=3x^2$ and we set $3x^2=0$, for which $x=0$ is the solution—but we can easily see from the graph on the right that $x=0$ is neither a minimum nor a maximum: it’s what we call an inflection point. To define inflection points, we’ll need to resort to our second-favorite tool, after the first derivative: the second derivative!

Taking the second derivative of $h$ (Yes, purists, I should’ve said $h$ with respect to $x$. Good work; I’m proud of you.), $h”(x)=6x$ and we notice that $h”(0)=0$ as well as $h'(0)=0$. Conversely, $f”(x)=\frac{4x^3}{25}-\frac{8x}{5}$ which is solved to find that $f”(x)=0$ for $x=\{0, \pm\sqrt{10}\}$. Doesn’t the part of the graph of $y=f(x)$ around $x=0$ look quite a lot like the same part of the graph of $y=h(x)=x^3$? (It does.) That too is an inflection point, and based on two pieces of evidence, we’ll suggest that an inflection point is a point $(x,f(x))$ on a function $f$ such that $f'(x)=0$ and $f”(x)=0$.

Actually, that’s not quite right, but it’s close. If we go through the function’s derivatives until we find one that’s nonzero, it has to be an odd-ordered derivative (i.e., the third derivative must be nonzero if the fourth is nonzero; if the third and fourth are zero, the fifth had better be nonzero if the sixth is too).

This brings us to the question of how we can determine analytically whether a point that has a zero derivative is a local minimum or maximum (we’ve already handled how to find if it’s neither). As the sharp ones (sharp as a thumbtack?) out there have guessed, that also involves the second derivative.

Let’s look back at $f”(x)$. We know that $x=-2$ and $x=4$ are local minima, and $f”(-2)=\frac{48}{25}$ while $f”(4)=\frac{96}{25}$. Similarly (oppositely?), $x=-4$ and $x=2$ are local maxima, and $f”(-4)=-\frac{96}{25}$ while $f”(2)=-\frac{48}{25}$.

Do you see the pattern? We’ll again draw a broad conclusion from a very little bit of evidence, but it’s true: for a function $f(x)$,

  • local minima are points where $f'(x)=0$ and $f”(x)>0$
  • local maxima are points where $f'(x)=0$ and $f”(x)<0$
  • stationary points are points where $f'(x)=0$ and $f”(x)=0$ and the lowest-order nonzero derivative has a order that’s an odd number

I know this is a long lesson—we’re almost done. Finding global extrema seems pretty easy: just compare all the local extrema. Right? Kinda. Some functions (like $\sin x$) have no global extrema, just a lot of equal local ones. Also, the AP test is fond of asking about extrema on an interval, in which case you’ve also got to test the function’s value at each end of the interval. Okay, now I’m done.

Valid XHTML 1.1!
Valid CSS!