Maximums and minimums

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We use derivatives to help locate extrema.

Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function looks like. We can obtain a good picture of the graph using certain crucial information provided by derivatives of the function.

Extrema

Local extrema on a function are points on the graph where the $y$ -coordinate is larger (or smaller) than all other $y$ -coordinates on the graph at points ‘‘close to’’ $(x,y)$ .

(a): A function $f$ has a local maximum at $x=a$ , if $f(a)\ge f(x)$ for every $x$ near $a$ .
(b): A function $f$ has a local minimum at $x=a$ , if $f(a)\le f(x)$ for every $x$ near $a$ .

A local extremum is either a local maximum or a local minimum.

Local maximum and minimum points are quite distinctive on the graph of a function, and are therefore useful in understanding the shape of the graph. In many applied problems we want to find the largest or smallest value that a function achieves (for example, we might want to find the minimum cost at which some task can be performed) and so identifying maximum and minimum points will be useful for applied problems as well.

Critical points

If $(x,f(x))$ is a point where $f$ reaches a local maximum or minimum, and if the derivative of $f$ exists at $x$ , then the graph has a tangent line and the tangent line must be horizontal. This is important enough to state as a theorem, though we will not prove it.

Fermat’s Theorem If $f$ has a local extremum at $x=a$ and $f$ is differentiable at $a$ , then $f'(a)=0$ .

Does Fermat’s Theorem say that if $f'(a) = 0$ , then $f$ has a local extrema at $x=a$ ?

yes no

Consider $f(x) = x^3$ , $f'(0) = 0$ , but $f$ does not have a local maximum or minimum at $x=0$ .

Fermat’s Theorem says that the only points at which a function can have a local maximum or minimum are points at which the derivative is zero, consider the plots of $f(x) = x^3-4x^2+3x$ and $f'(x) = 3x^2-8x+3$ ,

or the derivative is undefined, as in the plot of $f(x) = x^{2/3}$ and $f'(x) = \frac {2}{3x^{1/3}}$ :

This brings us to our next definition.

A function has a critical point at $x=a$ if $f'(a) = 0\qquad \text {or}\qquad \text {$f'(a)$ does not exist.}$

When looking for local maximum and minimum points, you are likely to make two sorts of mistakes:

You may forget that a maximum or minimum can occur where the derivative does not exist, and so forget to check whether the derivative exists everywhere.
You might assume that any place that the derivative is zero is a local maximum or minimum point, but this is not true, consider the plots of $f(x) = x^3$ and $f'(x) = 3x^2$ .

While $f'(0)=0$ , there is neither a maximum nor minimum at $(0,f(0))$ .

Since the derivative is zero or undefined at both local maximum and local minimum points, we need a way to determine which, if either, actually occurs. The most elementary approach is to test directly whether the $y$ coordinates near the potential maximum or minimum are above or below the $y$ coordinate at the point of interest.

It is not always easy to compute the value of a function at a particular point. The task is made easier by the availability of calculators and computers, but they have their own drawbacks: they do not always allow us to distinguish between values that are very close together. Nevertheless, because this method is conceptually simple and sometimes easy to perform, you should always consider it.

Find all local maximum and minimum points for the function $f(x)=x^3-x$ .

Write $\ddx f(x)=\answer [given]{3x^2-1}.$ This is defined everywhere and is zero at $x=\pm \sqrt {3}/3$ . Looking first at $x=\sqrt {3}/3$ , we see that $f(\sqrt {3}/3)=\answer [given]{-2\sqrt {3}/9}.$ Now we test two points on either side of $x=\sqrt {3}/3$ , making sure that neither is farther away than the nearest critical point; since $\sqrt {3}<3$ , $\sqrt {3}/3<1$ and we can use $x=0$ and $x=1$ . Since $f(0)=0>-2\sqrt {3}/9\qquad \text {and}\qquad f(1)=0>-2\sqrt {3}/9,$ there must be a local minimum at $x=\answer [given]{\sqrt {3}/3}$ .

For $x=-\sqrt {3}/3$ , we see that $f(-\sqrt {3}/3)=2\sqrt {3}/9$ . This time we can use $x=0$ and $x=-1$ , and we find that $f(-1)=f(0)=0< 2\sqrt {3}/9$ , so there must be a local maximum at $x=\answer [given]{-\sqrt {3}/3}$ , see the plot below:

The first derivative test

The method of the previous section for deciding whether there is a local maximum or minimum at a critical point by testing ‘‘near-by’’ points is not always convenient. Instead, since we have already had to compute the derivative to find the critical points, we can use information about the derivative to decide. Recall that

If $f'(x) >0$ on an interval, then $f$ is increasing on that interval.
If $f'(x) <0$ on an interval, then $f$ is decreasing on that interval.

So how exactly does the derivative tell us whether there is a maximum, minimum, or neither at a point? Use the first derivative test.

First Derivative Test Suppose that $f$ is continuous on an interval, and that $f'(a)=0$ for some value of $a$ in that interval.

If $f'(x)>0$ to the left of $a$ and $f'(x)<0$ to the right of $a$ , then $f(a)$ is a local maximum.
If $f'(x)<0$ to the left of $a$ and $f'(x)>0$ to the right of $a$ , then $f(a)$ is a local minimum.
If $f'(x)$ has the same sign to the left and right of $a$ , then $f(a)$ is not a local extremum.

Consider the function $f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2$ Find the intervals on which $f$ is increasing and decreasing and identify the local extrema of $f$ .

Start by computing $\ddx f(x) = \answer [given]{x^3+x^2-2x}.$ Now we need to find when this function is positive and when it is negative. To do this, solve $f'(x) = \answer [given]{x^3+x^2-2x} =0.$ Factor $f'(x)$ $\begin{align*} f'(x) &= \answer[given]{x^3+x^2-2x} \\ &=x(\answer[given]{x^2+x-2})\\ &=x(x+2)\answer[given]{(x-1)}. \end{align*}$

So the critical points (when $f'(x)=0$ ) are when $x=-2$ , $x=0$ , and $x=1$ . Now we can check points between the critical points to find when $f'(x)$ is increasing and decreasing:

$\begin{align*} f'(-3)&=\answer[given]{-12},\\ f'(.5)&=\answer[given]{-0.625},\\ f'(-1)&=\answer[given]{2},\\ f'(2)&=\answer[given]{8}. \end{align*}$

From this we can make a sign table:

Hence $f$ is increasing on $(-2,0)$ and $(1,\infty )$ and $f$ is decreasing on $(-\infty ,-2)$ and $(0,1)$ . Moreover, from the first derivative test, the local maximum is at $x=0$ while the local minima are at $x=-2$ and $x=1$ , see the graphs of $f(x) =x^4/4 + x^3/3 -x^2$ and $f'(x) = x^3 + x^2 -2x$ .

Hence we have seen that if $f'$ is zero and increasing at a point, then $f$ has a local minimum at the point. If $f'$ is zero and decreasing at a point then $f$ has a local maximum at the point. Thus, we see that we can gain information about $f$ by studying how $f'$ changes. This leads us to our next section.