Finding Maximums and Minimums

We can try to find where a function $f$ is largest or smallest by evaluating $f$ at lots of values of $x$, a method which is not very efficient and may not find the exact place where $f$ achieves its extreme value. However, if we try hundreds or thousands of values for $\mathrm{x}$, then we can often find a value of $\mathrm{f}$ which is close to the maximum or minimum. In general, this type of exhaustive search is only practical if you have a computer do the work.

The graph of a function is a visual way of examining lots of values of $f$, and it is a good method, particularly if you have a computer to do the work for you. However, it is inefficient, and we still may not find the exact location of the maximum or minimum.

Calculus provides ways of drastically narrowing the number of points we need to examine to find the exact locations of maximums and minimums. Instead of examining $\mathrm{f}$ at thousands of values of $\mathrm{x}$, calculus can often guarantee that the maximum or minimum must occur at one of 3 or 4 values of $x$, a substantial improvement in efficiency.

Before we examine how calculus can help us find maximums and minimums, we need to define the concepts we will develop and use.

Fig. 1

Global and local minimums are defined similarly by replacing the $\geq$ with $\leq$ in the previous definitions.

The local and global extremes of the function in Fig. 2 are labeled. You should notice that every global extreme is also a local extreme, but there are local extremes which are not global extremes. If $\mathrm{h}(\mathrm{x})$ is the height of the earth above sea level at the location $\mathrm{x}$, then the global maximum of $\mathrm{h}$ is $\mathrm{h}$ (summit of Mt. Everest) =29,028 feet. The local maximum of $\mathrm{h}$ for the United States is $\mathrm{h}$ (summit of Mt. McKinley)=20,320 feet. The local minimum of $\mathrm{h}$ for the United States is $\mathrm{h}$ (Death Valley)=-282 feet.

Practice 1: The table shows the annual calculus enrollments at a large university. Which years had relative maximum or minimum calculus enrollments? What were the global maximum and minimum enrollments in calculus?

$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c} \text { year } & 1980 & 81 & 82 & 83 & 84 & 85 & 86 & 87 & 88 & 89 & 90 \\ \hline \text { enrollment } & 1257 & 1324 & 1378 & 1336 & 1389 & 1450 & 1523 & 1582 & 1567 & 1545 & 1571 \end{array}$

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.

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.

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 $\mathrm{x}$.

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

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?

So far we have been discussing finding extreme values of functions over the entire real number line or on an open interval, but, in practice, we may need to find the extreme of a function over some closed interval [$c, d$]. If the extreme value of $\mathrm{f}$ occurs at $\mathrm{x}=\mathrm{a}$ between $\mathrm{c}$ and $\mathrm{d}, \mathrm{c} < \mathrm{a} < \mathrm{d}$, then the previous reasoning and results still apply: either $\mathrm{f}^{\prime}(\mathrm{a})=0$ or $\mathrm{f}$ is not differentiable at $a$.

On a closed interval, however, there is one more possibility: an extreme can occur at an endpoint of the closed interval (Fig. 7), at $\mathrm{x}=\mathrm{c}$ or $\mathrm{x}=\mathrm{d}$.

Fig. 7

Practice 4: List all of the local extremes $(\mathrm{a}, \mathrm{f}(\mathrm{a}))$ of the function in Fig. 8 on the interval $[1,4]$ and state whether (i) $\mathrm{f}^{\prime}(\mathrm{a})=0$ or (ii) $\mathrm{f}$ is not differentiable at a or (iii) a is an endpoint.

Fig. 8

Example 3: Find the extreme values of $f(x)=x^{3}-3 x^{2}-9 x+5$ for $-2 \leq x \leq 6$.

Solution: $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-6 \mathrm{x}-9=3(\mathrm{x}+1)(\mathrm{x}-3)$. We need to find where (i) $\mathrm{f}^{\prime}(\mathrm{x})=0$, (ii) $\mathrm{f}$ is not differentiable, and (iii) the endpoints.

Sometimes the function we need to maximize or minimize is more complicated, but the same methods work.

Example 4: Find the extreme values of $f(x)=\frac{1}{3} \sqrt{64+x^{2}}+\frac{1}{5}(10-x)$ for $0 \leq x \leq 10$.

Solution: This function comes from an application we will examine in section 3.5. The only possible locations of extremes are where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ is undefined or where $x$ is an endpoint of the interval $[0,10]$.

$\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{D}\left(\frac{1}{3}\left(64+\mathrm{x}^{2}\right)^{1 / 2}+\frac{1}{5}(10-\mathrm{x})\right)=\frac{1}{3} \frac{1}{2}\left(64+\mathrm{x}^{2}\right)^{-1 / 2}(2 \mathrm{x})-\frac{1}{5}=\frac{\mathrm{x}}{3 \sqrt{64+\mathrm{x}^{2}}}-\frac{1}{5}$

To determine where $f^{\prime}(x)=0$, we need to set the derivative equal to $0$ and solve for $x$.

If $f^{\prime}(x)=\frac{x}{3 \sqrt{64+x^{2}}}-\frac{1}{5}=0$ then $\frac{x}{3 \sqrt{64+x^{2}}}=\frac{1}{5}$ so $\frac{x^{2}}{576+9 x^{2}}=\frac{1}{25}$

Then $16 \mathrm{x}^{2}=576$ so $\mathrm{x}=\pm 6$, and the only point  in the interval $[0,10]$ where $\mathrm{f}^{\prime}(\mathrm{x})=0$ is at $\mathrm{x}=6$.

Putting $\mathrm{x}=6$ into the original equation for $\mathrm{f}$ gives $\mathrm{f}(6) \approx 4.13$.

We can evaluate the formula for $\mathrm{f}^{\prime}(\mathrm{x})$ for any value of $\mathrm{x}$, so the derivative is always defined. Finally, the interval $[0,10]$ has two endpoints, $x=0$ and $x=10$. $f(0) \approx 4.67$ and $f(10) \approx 4.27$.

The maximum of $\mathrm{f}$ on $[0,10]$ must occur at one of the points $(0,4.67)$, $(6,4.13)$ and $(10,4.27)$, and the minimum must occur at one of these three points.

The maximum value of $\mathrm{f}$ is $4.67$ at $\mathrm{x}=0$, and the minimum value of $\mathrm{f}$ is $4.13$ at $\mathrm{x}=6$. The graph of $\mathrm{f}$ is shown in Fig. 9.

Fig. 9

Practice 5: Find the extreme values of $f(x)=\frac{1}{3} \sqrt{64+x^{2}}+\frac{1}{5}(10-x)$ for $0 \leq x \leq 5$.

The points at which a function might have an extreme value are called critical numbers.

If we are trying to find the extreme values of $\mathrm{f}$ on an open interval $\mathrm{c} < \mathrm{x} < \mathrm{d}$ or on the entire number line, then there will not be any endpoints so there will not be any endpoint critical numbers to worry about.

We can now give a very succinct description of where to look for extreme values of a function:

The critical numbers only give the possible locations of extremes, and some critical numbers are not the locations of extremes. The critical numbers are the candidates for the locations of maximums and minimums (Fig. 10). Section 3.5 is devoted entirely to translating and solving maximum and minimum problems.

Fig. 10

Which Functions Have Extremes?

So far we have concentrated on finding the extreme values of functions, but some functions don't have extreme values. Example 2 showed that $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}$ did not have a maximum or minimum.

Example 5: Find the extreme values of $\mathrm{f}(\mathrm{x})=\mathrm{x}$.

Solution: Since $\mathrm{f}^{\prime}(\mathrm{x})=1 > 0$ for all $\mathrm{x}$, the first theorem in this section guarantees that $\mathrm{f}$ has no extreme values. The function $\mathrm{f}(\mathrm{x})=\mathrm{x}$ does not have a maximum or minimum on the real number line.

The difficulty with the previous function was that the domain was so large that we could always make the function larger or smaller than any given value. The next example shows that we can encounter the same difficulty even on a small interval.

Example 6: Show that $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}$ (Fig. 11) does not have a maximum or minimum on the interval $(0,1)$.

Fig. 11

Solution: $\mathrm{f}$ is continuous for all $\mathrm{x} \neq 0$ so $\mathrm{f}$ is continuous on the interval $(0,1)$ For $0 < x < 1$, $f(x)=\frac{1}{x} > 0$ (Fig.11). For any number a strictly between $0$ and $1$, we can show that $\mathrm{f}(\mathrm{a})$ is neither a maximum nor a minimum of $\mathrm{f}$ on $(0,1)$. Pick $b$ to be any number between $0$ and $a$, $0 < b < a$. Then $\mathrm{f}(\mathrm{b})=\frac{1}{\mathrm{~b}} > \frac{1}{\mathrm{a}}=\mathrm{f}(\mathrm{a})$, so $\mathrm{f}(\mathrm{a})$ is not a maximum. Similarly, pick $\mathrm{c}$ to be any number between $a$ and $1$, $\mathrm{a} < \mathrm{c} < 1$. Then $\mathrm{f}(\mathrm{a})=\frac{1}{\mathrm{a}} > \frac{1}{\mathrm{c}} = \mathrm{f}(\mathrm{c})$, so $\mathrm{f}(\mathrm{a})$ is not a minimum. The interval $(0,1)$ is not large, but $\mathrm{f}$ still does not have an extreme value in $(0,1)$.

The Extreme Value Theorem gives conditions so that a function is guaranteed to have a maximum and a minimum.

The proof of this theorem is difficult and is omitted. Fig. 12 illustrates some of the possibilities for continuous and discontinuous functions on open and closed intervals. The Extreme Value Theorem guarantees that certain functions (continuous) on certain intervals (closed) must have maximums and minimums. Other functions on other intervals may or may not have maximums and minimums.

Fig. 12

Practice 1: The enrollments were relative maximums in $'82$, $'87$, and $'90$.
The global maximum was in $'87$. The enrollments were relative minimums in $'80$, $'83$, and $'89$ . The global minimum occurred in $'80$.

Practice 2: $f(x)=x^{2}+4 x-5$ is a polynomial so $f$ is differentiable for all $x$, and $f^{\prime}(x)=2 x+4$. $f^{\prime}(x)=0$ when $x=-2$ so the only candidate for a local extreme is $x=-2$. Since the graph of $f$ is a parabola opening up, the point $(-2, f(-2))=(-2,-9)$ is a local minimum.

$g(x)=2 x^{3}-12 x^{2}+7$ is a polynomial so $g$ is differentiable for all $x$, and $g^{\prime}(x)=6 x^{2}-24 x=6 x(x-4)$. $g^{\prime}(x)=0$ when $x=0,4$ so the only candidates for a local extreme are $x=0$ and $x=4$. The graph of $g$ (Fig. 22) shows that $g$ has a local maximum at $(0,7)$ and a local minimum at $(4,-57)$.

Fig. 22

Practice 3:

$\begin{array}{l|l|l|l}x & f(x) & f^{\prime}(x) & \max / \mathrm{min} \\\hline 1 & 5 & 0 & \text { local max } \\2 & & & \\3 & 1 & 0 & \text { local min } \\4 & 3 & 0 & \text { neither } \\5 & & & \\6 & 7 & 0 & \text { local max }\end{array}$

see Fig. 23

Fig. 23

Practice 4: (1, $\mathrm{f}(1)$) is a local minimum. $\mathrm{x}=1$ is an endpoint.

(2, $f(2)$) is a local maximum. $f^{\prime}(2)=0$

(3, $f(3)$) is a local minimum. $f$ is not differentiable at $x=3$.

(4, $f(4)$) is a local maximum. $x=4$ is an endpoint.

Critical points: endpoints $\mathrm{x}=0$ and $\mathrm{x}=\mathbf{5}$.

$\mathrm{f}$ is differentiable for all $0 < \mathrm{x} < 5$: none.

$\mathrm{f}^{\prime}(\mathrm{x})=0$: none in $[0,5]$.

$f(0) \approx 4.67$ is the maximum of $\mathrm{f}$ on $[0,5]$. $\mathrm{f}(5) \approx 4.14$ is the minimum of $\mathrm{f}$ on $[0,5]$