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Math 112 Solutions for 5.6, p. 504

1.
(b) $ \frac{2}{11}$. (c) $ \frac{6}{5}(2^{5/2}-1)$. (d) $ \frac{381}{7}$. (f) $ \frac{7}{16}$. (h) $ 3(2^{1/3}-1)$. (j) $ \frac{2001}{32}$. (l) 6.

2.
(b) $ \frac{1}{2}$. (c) -4. (d) $ \sqrt 2-1$. (f) $ -\frac{2}{\sqrt
3}+\sqrt 2$. (g) 2. (h) $ -1+\sqrt 3$. (j) $ \frac{5}{2}\sqrt 2-1$. (k) 1. (l) $ -\frac{1}{3}$.

3.
(b) $ \frac{2}{\ln 3}$. (c) $ 1-\frac{1}{e}$. (d) $ \frac{\sqrt
2-1}{\sqrt 2\ln 2}$. (f) $ \ln\frac{5}{2}$. (g) $ frac{\pi}{4}$. (h) $ \frac{\pi}{6}$. (j) $ 2\ln
\frac{8}{5}-0.171$. (l) $ \cosh 1-2\sinh 1-1$.

4.
(b) $ x^{3}+x^{-1}$. (c) $ \cos 2x$. (d) $ \dfrac{1}{x^{2}+1}$

5.
Let $ u=g(x)$. Then

$\displaystyle \frac{d}{dx}\left(\int_{a}^{g(x)}f(t)dt\right)=
\frac{d}{du}\left(\int_{a}^{u}f(t)dt\right)\frac{du}{dx}=f(u)\frac{du}{dx}
=f(g(x))g'(x).
$

6.
(b) $ \dfrac{1}{2\sqrt x\,(\sqrt x+2)}$. (c) $ 26x^{2}+8$.

(b) $ s(T)=\int_{0}^{T}r(t)dt$.

(c) (i) $ a(T)=\frac{s(T)}{T}=\frac{1}{T}\int_{0}^{T}r(t)dt$. (ii) $ a(T)$ is the slope of the line from the origin to the point $ (T,s(T))$.

(d) (i) $ s(T)$ has maximum slope at about $ T=5.5$ (ii) If $ a(T)$ is maximum, then $ a'(T)=0\Rightarrow
\frac{s'(T)T-s(T)}{t^{2}}=0\Rightarrow s'(T)=\frac{s(T)}{T}=a(T)$. (iii) at about 5.5 weeks. (iv) Since $ s'(T)=r(T)$, the condition is $ r(T)=a(T)$. (v) When the current sales level $ r$ equals the running average $ a$, then $ a$ is maximum. Beyond that, $ r$ is decreasing, which means that $ a$ will also be decreasing.

11.
If $ f(x)=A'(x)$ is increasing, $ A(x)$ is concave upward. Similarly, if $ f(x)=A'(x)$ is decreasing, then $ A(x)$ is concave downward.




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Jason Grout 2003-04-04