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Math 112 Solutions for 3.3, p. 248

(b) global minimum at $ (-1,2)$. (f) global maximum at $ \left(\dfrac{1}{\sqrt[3]{2}},\dfrac{1}{4}\right)$. (j) global maximum at $ (3,\frac{8}{3})$. (n) global minimum at the origin.
(b) global maximum at $ \left(\dfrac{1}{\sqrt 2},\sqrt 2\right)$, global minimum at $ (-1,-1)$, and local minimum at $ (1,1)$.

(c) global maximum of $ \sqrt 2$ at $ \frac{\pi}{4}+2n\pi$, global minimum of $ -\sqrt 2$ at $ \frac{5\pi}{4}+2n\pi, n\in \mathbb{Z}$.

(f) local maximum at $ (-\frac{\pi}{3}+2n\pi,-\frac{\pi}{3}+2n\pi+\sqrt 3)$, local minimum at $ (\frac{\pi}{3}+2n\pi,\frac{\pi}{3}+2n\pi-\sqrt 3)$.

(j) global minimum at $ (\frac{1}{3},-\frac{2}{3e})$.

(n) no extrema

The function $ f(x)=\begin{cases}x-a, &a\leq x<\frac{1}{2}(a+b)\\
-\frac{1}{2}(x-b), &\frac{1}{2}(a+b)\leq x \leq b \end{cases}$ Is defined on $ [a,b]$ but has no maximum there.

$ f''$ changes sign at $ c$, so $ (f')'$ changes sign at $ c$ and hence $ f'$ has an extremum.

Theorem. If $ f''(c)=0$ and $ f'''(c)\neq 0$, then $ f$ has an inflection point at $ c$.

Proof. If $ f''(c)=0$, then $ c$ is a critical point of $ f'$ at which the derivative is zero, so we may apply the second derivative test to $ f'$. If $ (f')''(c)\neq 0$, then $ f'$ has an extremum at $ c$, so that the derivative of $ f'$ changes at $ c$ and hence the concavity changes at $ c$. Thus $ c$ is an inflection point of $ f$.

Since $ f$ is continuous on a closed interval, it has a global maximum there, which is also a local maximum, so occurs at $ c$.

Suppose $ f$ is defined in $ (a,b)$ and $ c\in (a,b)$ is the only critical point of $ f$, and $ f$ has a local maximum at $ c$ (the case of minimum is similar). Since $ f$ is continuous and $ c$ is the only critical point, $ f$ is differentiable in $ (a,b)$ except possibly at $ c$. If $ f$ were to have a value greater than $ f(c)$ somewhere, then it must have the same value at some point, and then Rolle's Theorem would give us another critical point, which we can't have. Thus the value at $ c$ is a global maximum.

(a) P2 should maximize $ S$ with respect to $ y$, getting $ y=\dfrac{27-x}{4}$.

(b) P1 should replace $ y$ in $ S$ with the value that P2 is going to use, and minimize the resulting $ S$, getting $ x=-1$.

(c) P1 should minimize $ S$ with respect to $ x$, getting $ x=\dfrac{y-9}{2}$.

(d) P2 should replace $ x$ in $ S$ with the value that P1 is going to use and maximize the resulting $ S$, getting $ y=7$.

(e) Each player has to ``go first" so P1 chooses $ x=-1$ and P2 chooses $ y=7$. (The resulting score is 0, so the game is uninteresting, even though it is fair.)

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Jason Grout 2003-02-25