Finding Maximums and Minimums

Read this section to learn about maximums, minimums, and extreme values for functions. Work through practice problems 1-5.

Finding Maximums and Minimums of a Function

One way to narrow our search for a maximum value of a function $f$ is to eliminate those values of $x$ which, for some reason, cannot possibly make $f$ maximum.

Theorem: If $\mathrm{f}'(\mathrm{a}) > 0$ or $\mathrm{f}'(\mathrm{a}) < 0$ ,
then $\mathrm{f}(\mathrm{a})$ is not a local maximum or minimum. (Fig. 3)

Fig. 3

Proof: Assume that $\mathrm{f}^{\prime}(\mathrm{a})>0$ . By definition, $\mathrm{f}^{\prime}(\mathrm{a})=\lim\limits _{\Delta x \rightarrow 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}$ , so $\lim \limits _{\Delta x \rightarrow 0} \frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$ and the right and left limits are both positive: $\lim\limits _{\Delta x \rightarrow 0^{+}} \frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$ and $\quad \lim\limits _{\Delta x \rightarrow 0^{-}} \frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$ .

Since the right limit, $\Delta x \rightarrow 0^{+}$ , is positive, there are values of $\Delta x > 0$ so $\frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$ .

Multiplying each side of this last inequality by the positive $\Delta x$ , we have $f(a+\Delta x)-f(a) > 0$ and $f(a+\Delta x)>f(a)$ so $f(a)$ is not a maximum.

Since the left limit, $\Delta x \rightarrow 0^{-}$ , is positive, there are values of $\Delta x < 0$ so $\frac{f(a+\Delta x)-f(a)}{\Delta x} > 0$ .

Multiplying each side of the last inequality by the negative $\Delta x$ , we have that $f(a+\Delta x)-f(a) < 0$ and $f(a+\Delta x) < f(a)$ so $f(a)$ is not a minimum.'

The proof for the $"f '(a) < 0 "$ case is similar.

When we evaluate the derivative of a function

$f$ at a point

$x=a$ , there are only four possible outcomes:

$\mathrm{f}^{\prime}(\mathrm{a}) > 0, \mathrm{f}^{\prime}(\mathrm{a}) < 0, \mathrm{f}^{\prime}(\mathrm{a})=0$ or

$\mathrm{f}^{\prime}(\mathrm{a})$ is undefined. If we are looking for extreme values of

$\mathrm{f}$ , then we can eliminate those points at which

$\mathrm{f}'$ is positive or negative, and only two possibilities remain:

$\mathrm{f}^{\prime}(\mathrm{a})=0$ or

$\mathrm{f}$

$'(a)$ is undefined.

Theorem: If

$\mathrm{f}$ is defined on an open interval, and

$\mathrm{f}(\mathrm{a})$ is a local extreme of

$\mathrm{f}$ , then either

$\mathrm{f}^{\prime}(\mathrm{a})=0$ or

$\mathrm{f}$ is not differentiable at

$a$ .

Example 1: Find the local extremes of $f(x)=x^{3}-6 x^{2}+9 x+2$ for all values of $x$ .

Solution: An extreme value of $f$ can occur only where $f^{\prime}(x)=0$ or where $f$ is not differentiable. $f^{\prime}(x)=3 x^{2}-12 x+9=3\left(x^{2}-4 x+3\right)=3(x-1)(x-3)$ so $f^{\prime}(x)=0$ only at $x=1$ and $x=3 . f^{\prime}$ is a polynomial, so $\mathrm{f}$ is differentiable for all $\mathrm{x}$ .

The only possible locations of local extremes of $\mathrm{f}$ are at $\mathrm{x}=1$ and $\mathrm{x}=3$ . We don't know yet whether $\mathrm{f}(1)$ or $f(3)$ is a local extreme of $f$ , but we can be certain that no other point is a local extreme. The graph of $\mathrm{f}$ (Fig. 4 ) shows that $(1, f(1))=(1,6)$ is a local maximum and $(3, \mathrm{f}(3))=(3,2)$ is a local minimum. This function does not have a global maximum or minimum.

Fig. 4

Practice 2: Find the local extremes of $f(x)=x^{2}+4 x-5$ and $g(x)=2 x^{3}-12 x^{2}+7$ .

It is important to recognize that the conditions " $\mathrm{f}^{\prime}(\mathrm{a})=0$ " or " $\mathrm{f}$ not differentiable at a " do not guarantee that $\mathrm{f}(\mathrm{a})$ is a local maximum or minimum. They only say that $\mathrm{f}(\mathrm{a})$ might be a local extreme or that $\mathrm{f}(\mathrm{a})$ is a candidate for being a local extreme.

Example 2: Find all local extremes of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ .

Solution: $f(x)=x^{3}$ is differentiable for all $x$ , and $f^{\prime}(x)=3 x^{2}$ . The only place where

$f^{\prime}(x)=0$ is at $x=0$ , so the only candidate is the point $(0,0)$ . But if $x > 0$ then

$f(x)=x^{3} > 0=f(0)$ , so $f(0)$ is not a local maximum. Similarly, if $x < 0$ then

$f(x)=x^{3} < 0=f(0)$ so $f(0)$ is not a local minimum. The point $(0,0)$ is the only candidate to be a local extreme of $\mathrm{f}$ , and this candidate did not turn out to be a local extreme of $f$ . The function $f(x)=x^{3}$ does not have any local extremes. (Fig. 5 )

Fig. 5

If $\mathrm{f}^{\prime}(\mathrm{a})=0$ or $\mathrm{f}$ is not differentiable at $a$

then the point $(a, f(a))$ is a candidate to be a local extreme and may or may not be a local extreme.

Practice 3: Sketch the graph of a differentiable function $\mathrm{f}$ which satisfies the conditions:
(i) $f(1)=5, f(3)=1, f(4)=3$ and $f(6)=7$ ,
(ii) $\mathrm{f}^{\prime}(1)=0, \mathrm{f}^{\prime}(3)=0, \mathrm{f}^{\prime}(4)=0$ and $\mathrm{f}^{\prime}(6)=0$ ,
(iii) the only local maximums of $f$ are at $(1,5)$ and $(6,7)$ , and the only local minimum is at $(3,1)$ .

Is f(a) a Maximum or Minimum or Neither?

Once we have found the candidates $(a, f(a))$ for extreme points of $f$ , we still have the problem of determining whether the point is a maximum, a minimum or neither.

One method is to graph (or have your calculator graph) the function near a, and then draw your conclusion from the graph. All of the graphs in Fig. 6 have $\mathrm{f}(2)= 3$ , and, on each of the graphs, $\mathrm{f} '(2)$ either equals $0$ or is undefined. It is clear from the graphs that the point $(2,3)$ is a local maximum in (a) and (d), $(2,3)$ is a local minimum in (b) and (e), and $(2,3)$ is not a local extreme in (c) and (f).

Fig. 6

In sections 3.3 and 3.4, we will investigate how information about the first and second derivatives of $\mathrm{f}$ can help determine whether the candidate $(\mathrm{a}, \mathrm{f}(\mathrm{a}))$ is a maximum, a minimum, or neither.