Finding Maximums and Minimums

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 $[0,5]$