MA005: Calculus I

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 $0 \leq x \leq 5$.