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Math 112 Solutions for 3.2, p. 237

1.
(a) none. (c) 0, $ \frac{3}{4}$. (d) $ -\dfrac{1}{2^{1/3}}$. (e) o, 3, -3. (g) 1. (i) $ \frac{\pi}{4}+n\pi, n\in
\mathbb{Z}$. (k) $ \pi +2n\pi, n\in
\mathbb{Z}$. (o) 0.
2.
(a) increasing and concave upward. (b) concave downward. (c) decreasing and concave upward. (d) increasing and concave downward. (e) decreasing and concave downward.
6.
Since $ f$ is of odd degree, it has at least one real zero. If $ f$ has two real zeros, then $ f$ has a critical point, by Rolle's Theorem. But $ f '(x)=3x^{2}+4 >0$ for every $ x$, and $ f$ has no critical point, so at most one real zero.
8.
(c) $ \tan 0\neq \tan \pi/3$
9.
Let $ f(t),g(t)$ be the positions of the cars, and we assume $ f$ and $ g$ are differentiable. Let $ h(t)=f(t)-g(t)$. Since each car passes the other, there are two times $ t_{1}$ and $ t_{2}$ at which they have the same position, so $ h(t_{1})=h(t_{2})=0$. Hence by Rolle's Theorem there is a time $ T$ between $ t_{1}$ and $ t_{2}$ such that $ h'(T)=0$, so that $ f '(T)=g'(T)$ and the cars have the same speed at time $ T$.
10.
Since the cars pass each other, there is a time $ T_{1}$ at which their speeds are the same; when they pass each other again, there is a time $ T_{2}$ at which their speeds are the same. Hence there is a time between $ T_{1}$ and $ T_{2}$ when the derivatives of their speeds, their accelerations, are the same.
11.
(a) $ c=1$ (c) $ c=\ln(\frac{e^e-1}{e})\approx 1.65002$ (d) $ c=\dfrac{1}{\ln(1+e)}\approx 0.31326$.
12.
(d) $ x^{2/3}$ is not differentiable at 0.
13.
(c) cp: 0, 2. hcp: 1. increasing in $ (-\infty,0)$ and in $ (2,\infty)$. decreasing in $ (0,2)$. concave down in $ (-\infty,1)$. concave up in $ (1,\infty)$. infl.pt. at 1

(g) cp: -1, 1. hcp: none (0 is not in the domain). increasing in $ (-\infty,-1)$ and in $ (1,\infty)$. decreasing in (-1, 0) and in (0, 1). concave down in $ (-\infty,0)$. concave up in $ (0,\infty)$. infl.pt.: none.

(h) cp: 0. increasing in $ (-\infty,0)$, decreasing in $ (0,\infty)$. hcp: $ \pm\frac{1}{\sqrt{3}}$. concave down in $ (\frac{-1}{\sqrt 3},\frac{1}{\sqrt
3})$, concave up in $ (-\infty, \frac{-1}{\sqrt 3})$ and in $ (\frac{1}{\sqrt
3},\infty)$.

(l) cp: none. increasing everywhere. hcp: $ n\pi$. concave down in $ (\pi+2n\pi,2\pi+2n\pi)$, concave up in $ (2n\pi,\pi+2n\pi)$.

(n) cp: 1. increasing in $ (-\infty,1)$, decreasing in $ (1,\infty)$. hcp: 2. concave down in $ (-\infty,2)$, concave up in $ (2,\infty)$.

(o) cp: 0. hcp: -1, 1. increasing in $ (-\infty,0)$. decreasing in $ (0,\infty)$. concave up in $ (-\infty,-1)$ and in $ (1,\infty)$. concave down in (-1, 1). infl.pt.: -1, 1.

14.
(c) vertical tangent line at 0.
15.
If $ f''(x)<0$, then $ f'$ is decreasing by Theorem 71, so $ f$ is concave downward.
16.
(a) $ y=e^{a}x+(1-a)e^{a}$. (b) $ (1-a)e^{a}$. (c) The curve is concave upward, so the tangent line, and hence its intercept, is always below the curve.
17.
(a) If $ p(c_{1})=p(c_{2})$, then by Rolle's Theorem there is a point between $ c_{1}$ and $ c_{2}$ at which $ p'(x)=0$. But then $ x$ is a critical point, a contradiction.
19.
(c) cusp. (d) vertical tangent line.
21.
(a) The curve is concave downward because as the temperature increases, the rate of temperature increase lessens. (b) $ 60\leq T(35)\leq 72$. (c) $ 68\leq T(35)\leq 72$. (d) between $ t=32.5$ and $ t=38.75$.

22.
Apply the Mean Value Theorem to $ f$ on the interval $ [a,x]$. Thus there exists $ c\in(a,x)$ such that $ f'(c)=\dfrac{f(x)-f(a)}{x-a}$, from which the result follows.




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Jason Grout 2003-02-21