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Math 112 Solutions for 3.7, p. 286

2.
$ f(x)=x^{2}-ax$ describes a parabola opening upward with zeros at 0 and $ a$ and vertex $ (\frac{a}{2},-\frac{a^{2}}{4})$. $ g(x)=ax-x^{2}$ has the same zeros but vertex $ (\frac{a}{2},\frac{a^{2}}{4})$ and opens downward.
3.
Let $ f(x)=ax^{4}+bx^{2}+c$. The critical points are 0 and $ \pm\sqrt{\frac{-b}{2a}}$. $ f''(0)=2b$ and $ f''(\pm\sqrt{\frac{-b}{2a}})=-4b$. Hence if $ a>0$ and $ b<0$, $ f$ has global minima at $ \pm\sqrt{\frac{-b}{2a}}$ and a local maximum at 0. If $ a>0$ and $ b>0$, $ f$ has only a global minimum at 0. If $ a<0$ and $ b>0$, $ f$ has global maxima at $ \pm\sqrt{\frac{-b}{2a}}$ and a local minimum at 0. If $ a<0$ and $ b<0$, $ f$ has only a global maximum at 0.

4.
Let $ f(x)=ax^{3}+bx^{2}+cx+d$. The critical points are at $ x=\dfrac{-b\pm\sqrt{b^{2}-3ac}}{3a}$ and the inflection point is at $ x=\dfrac{-b}{3a}$. The average of the $ x$-coordinates of the critical points is clearly the $ x$-coordinate of the inflection point. The same is true for the $ y$-coordinates.
5.
If the cubic in Problem 4 has only one critical point, it is because $ b^{2}-3ac<0$ and the critical point is $ x=\dfrac{-b}{3a}$, the inflection point.
8.
Let the cubic have the equation $ f(x)=ax^{3}+bx^{2}+cx+d$. If (3, 0) is the local maximum and (5, -1) is the inflection point, then the local minimum is at (7, -2) and $ a$ is positive. The inflection point is at $ x=\frac{-b}{3a}=5$, so $ b=-15a$. Since the first critical point is $ x=\dfrac{-b-\sqrt{b^{2}-3ac}}{3a}=3$, we get $ c=63a$. Finally, using $ f(3)=0$, we get $ d=-81a$. Hence $ f(x)=a(x^{3}-15x^{2}+63x-81)$ with $ a>0$.

10.
(a) Since 120 mi/hr = 176 ft/sec, the maximum steepness is $ \frac{8}{176}=\frac{1}{22}$, so the minimum slope is $ -\frac{1}{22}$.

(b) With $ f(x)=ax^{3}+bx^{2}+cx+d$, we let the inflection point be the origin; since the inflection point is at $ x=\frac{-b}{3a}$, we have $ b=0$. Since $ f(0)=0$, we have $ d=0$. The slope at the inflection point is $ f'(0)=c=\frac{-1}{22}$. The critical points are then at $ x=\pm\sqrt{\frac{-c}{3a}}=\pm\frac{1}{\sqrt{66a}}$ and $ y=\mp\frac{1}{33\sqrt{66a}}$. Therefore the run is $ \frac{2}{\sqrt{66a}}$ and the rise is $ -\frac{2}{33\sqrt{66a}}$.

(c) If the rise is -12,000 feet, then the run is $ \frac{2}{\sqrt{66a}}=(33)(12000)$ feet $ =75$ miles.

(d) From part (c) we have $ a=\frac{1}{66(33)^{2}(6000)^{2}}\approx 3.86\cdot
10^{-13}$.

11.
Increasing $ a$ magnifies the graph vertically. Increasing $ b$ compresses the graph horizontally.

13.
If the maximum of $ f(x)=axe^{-bx}$ is (7,55), then $ b=\frac{1}{7}$ and $ a=\frac{55e}{7}$. It follows that $ f(x)$ is less than 20 when $ \frac{55e}{7}xe^{-x/7}<20$, which is true for $ x>22.1422$.

14.
$ y=5[1-(\frac{4}{5})^{x}]$.

17.
$ \displaystyle{\lim_{x\to\infty}\frac{a}{1+be^{-kx}}=a\text{ and
} \lim_{x\to-\infty}\frac{a}{1+be^{-kx}}=0}$.

20.
Demand will be about 24 sales per week, and it will take another four weeks to reach it.

22.
about 1557.

25.
Think of $ y=e^{-ax}$ as an amplitude to a sunisoid.

26.
The graph is a bell curve having its maximum at $ x=\mu$ and inflection points at $ x=\mu\pm\sigma$.

27.
Let the chain hang from the points $ (-p,q)$ and $ (p,q)$ with its low point at the origin. Since the low point is a critical point, we get $ f'(0)=0=bc\sinh(d)$, so $ d=0$. Since the origin is on the curve, we have $ f(0)=0=a+b\cosh(0)=a+b$, and $ b=-a$. Then $ f(p)=q=a-a\cosh(cp)$, so $ a=\dfrac{q}{1-\cosh(cp)}$. Thus $ f(x)=\dfrac{q(1-\cosh cx)}{1-\cosh cp}$.




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Jason Grout 2003-03-06