[Latexpage]
Here are two self-tests. Yes, each one only has 3 problems.
answers and partial solutions below: I am still in the process of adding details
Self-test A
1a) to find y-intercept, set x = 0: but 0 is not in the domain of this function, so there is no y-intercept. To find x-intercepts, set f(x) = 0, which would mean 10 = 0, so there are no x-intercepts.
1b) There are vertical asymptotes at x=0 and x=3.
1c) $\displaystyle \lim_{x \rightarrow \infty}f(x) = 0$,and $\displaystyle \lim_{x \rightarrow -\infty}f(x) = 0$ also, so the horizontal asymptote is y=0, namely, the x-axis.
1d) f(x) is increasing on (1, 3) and decreasing on $(-\infty, 0) \cup (0, 1) \cup (3, \infty)$
1e) There is a local minimum at (1, 2.5) and no local maxima.
1f) Concave up on $(0, 3) \cup (3, \infty)$ and concave down on $(-\infty, 0)$
1g) There are no inflection points.
2) $V(x) = x^{3}$
$\frac{\mathrm{d}V}{\mathrm{d}t} = 3x^{2}\frac{\mathrm(d)x}{\mathrm{d}t} = 5$
so $\frac{\mathrm(d)x}{\mathrm{d}t} = \frac{5}{3x^{2}}$
The surface area is $A = 6x^{2}$
$\frac{\mathrm{d}A}{\mathrm{d}t} = 12x\frac{\mathrm(d)x}{\mathrm{d}t} = \frac{60}{3x^{2}}$
When x = 10, $\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{60}{3\cdot 10^{2}} = \frac{1}{15}$
3) The dimensions are: side parallel to river = 250nfeet, two other sides = 125 feet each
The maximum area is $250\cdot 125 = 31250$ square feet
Self-test B
1a) The x- and the y-intercept is (0,0)
1b) The vertical asymptotes are x = -2 and x = 2
1c) Find the limit of f(x) as x goes to $\infty$ and as x goes to $-\infty$: The horizontal asymptote is y = $-\frac{1}{2}$
1d) f(x) is increasing on $(-\infty, -2) \cup (-2, 0)$ and decreasing on $(0,2) \cup (2,\infty)$
1e) There is a local maximum at (0,0); no local minima
1f) The graph is concave up on $(-\infty, -2) \cup (2,\infty)$ and concave down on $(-2,2)$
1g) There are no inflection points
2) Volume of a cylinder with radius 5: $V = 25\pi h$
$\frac{\mathrm{d}V}{\mathrm{d}t} = 25\pi \frac{\mathrm{d}h}{\mathrm{d}t}$
Given that $\frac{\mathrm{d}V}{\mathrm{d}t} = 2$, solve for $\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{2}{25\pi}$
3) Surface area is $S = x^{2} + 4xy$
constraint: the volume must be 500 so $x^{2}y = 500 \implies y = \frac{500}{x^{2}}$
Substitute in the surface area:
$S = x^{2} + \frac{2000}{x}$
$S’ = 2x – \frac{2000}{x^{2}}$
The critical value is x = 10; check that it gives a local minimum, so the dimensions of the box are 10 by 10 by 5 inches
