Exam A Solutions

1. Find the derivative of $g(x)=\frac{x e^{x^{2}+1}}{3 x+2}$. Do not simplify your answer.

   Solution: Applying the quotient and product rules, we obtain $$\begin{aligned} g^{\prime}(x) & =\frac{(3 x+2) \frac{d}{d x}\left[x e^{x^{2}+1}\right]-x e^{x^{2}+1} \frac{d}{d x}[3 x+2]}{(3 x+2)^{2}} \\ & =\frac{(3 x+2)\left(x e^{x^{2}+1}(2 x)+e^{x^{2}+1}\right)-x e^{x^{2}+1}(3)}{(3 x+2)^{2}} \end{aligned}$$

2. If $f(9)=1, f^{\prime}(9)=3, g(1)=4, g^{\prime}(1)=-2$, and $h(x)=x^{3 / 2} g(f(x))$, then what is $h^{\prime}(9)$? Simplify your answer.

   Solution: Via the product and chain rules, we obtain $h^{\prime}(x)=x^{3 / 2} g^{\prime}(f(x)) f^{\prime}(x)+g(f(x))(3 / 2) x^{1 / 2}$. Thus $$\begin{aligned} h^{\prime}(9) & =9^{3 / 2} g^{\prime}(f(9)) f^{\prime}(9)+g(f(9))(3 / 2) 9^{1 / 2} \\ & =27 g^{\prime}(1)(3)+g(1)(3 / 2)(3) \\ & =27(-2)(3)+4(3 / 2)(3) \\ & =-144. \end{aligned}$$

3. Suppose $s(t)=t^{3}-9 t^{2}+15 t$ is the position of a particle traveling along a coordinate line, where $s$ is measured in meters and $t$, in seconds. At what times will the particle be at rest?

   Solution: The particle will be at rest when its velocity is zero; that is, when the rate of change in the particle’s position with respect to time is $0$. Thus we solve $\frac{d s}{d t}=0$: $$3 t^{2}-18 t+15=0 ; \quad \text{equivalently} \quad 3(t-1)(t-5)=0$$ Hence, the particle is at rest when $t=1$ and when $t=5$.

4. Find and classify the critical numbers of $f(x)=x^{5}(x-11)^{6}$, indicating for each critical number whether it yields a relative maximum value of $f$, a relative minimum value, or neither.

   Solution: We have $$\begin{aligned} f^{\prime}(x) & =x^{5} \cdot 6(x-11)^{5}+(x-11)^{6} \cdot 5 x^{4} \quad \text{(by the Product Rule)} \\ & =x^{4}(x-11)^{5}(6 x+5(x-11)) \\ & =x^{4}(x-11)^{5}(11 x-55) \\ & =11 x^{4}(x-11)^{5}(x-5). \end{aligned}$$ We see that the critical numbers of $f$ are $0$, $5$, and $11$, each of these being a zero of $f^{\prime}$.

   

   Original (raster)

   

   Reconstructed (TikZ) — RMSE: 0.2124 FAIL

   The $f^{\prime}$ sign-line above shows that $f$ is increasing on the intervals $(-\infty, 0)$, $(0,5)$, and $(11, \infty)$, while $f$ is decreasing on $(5,11)$. Because $f$ is continuous, we conclude via the first-derivative test that $0$ yields neither a relative maximum nor a relative minimum of $f$, that $5$ yields a relative maximum, and that $11$ yields a relative minimum.

5. Find $f(x)$ given that $f^{\prime}(x)=3 x^{2}+4 x-1$ and $f(2)=9$.

   Solution: Antidifferentiating both sides of the equation $f^{\prime}(x)=3 x^{2}+4 x-1$, we find $f(x)=x^{3}+2 x^{2}-x+C$ for some constant $C$. Now, using $f(2)=9$, we obtain $9=14+C$, so that $C=-5$. Thus $f(x)=x^{3}+2 x^{2}-x-5$.

6. Find the area of the region bounded by the curves $y=4-x^{2}$ and $y=0$.

   Solution: The region whose area is to be computed is shaded in blue below.

   

   Original (raster)

   

   Reconstructed (TikZ) — RMSE: 0.1181 PASS

   The definite integral $\int_{-2}^{2}\left(4-x^{2}\right) d x$ yields the area: $$\int_{-2}^{2}\left(4-x^{2}\right) d x=\left[4 x-\frac{x^{3}}{3}\right]_{-2}^{2}=(8-8 / 3)-(-8+8 / 3)=\frac{32}{3}.$$

7. The number of items produced by a manufacturer is given by $$p=100 x y^{3},$$ where $x$ is the amount of capital and $y$ is the amount of labor, amounts that change over time. At a particular point in time:

   1. the manufacturer has 2 units of capital;

   2. capital is increasing at a rate of 1 unit per month;

   3. the manufacturer has 3 units of labor; and

   4. labor is decreasing at a rate of $1 / 3$ unit per month.

   Determine the rate of change in the number of items produced at this point in time.

   Solution: Differentiating with respect to time we have $$\frac{d p}{d t}=100\left(x \cdot 3 y^{2} \frac{d y}{d t}+y^{3} \frac{d x}{d t}\right)$$ At the moment when $x=2, \frac{d x}{d t}=1, y=3$ and $\frac{d y}{d t}=-1 / 3$, we have that $$\frac{d p}{d t}=100\left(2 \cdot 3 \cdot 3^{2} \cdot\left(\frac{-1}{3}\right)+3^{3} \cdot 1\right)=900 \text { items } / \text { month. }$$

8. Find $\int\left(e^{x}+1\right)^{2} d x$.

   Solution: $$\begin{aligned} \int\left(e^{x}+1\right)^{2} d x & =\int\left(e^{2 x}+2 e^{x}+1\right) d x \\ & =\int e^{2 x} d x+2 \int e^{x} d x+\int 1 d x \\ & =\frac{1}{2} \int e^{u} d u+2 e^{x}+x+C \quad\left(u=2 x ; \frac{1}{2} d u=d x\right) \\ & =\frac{1}{2} e^{u}+2 e^{x}+x+C \\ & =\frac{1}{2} e^{2 x}+2 e^{x}+x+C \end{aligned}$$ Grade the problem "correct" even if you forgot to include $+C$.

9. Evaluate $\int_{e}^{e^{4}} \frac{1}{x \sqrt{\ln x}} d x$.

   Solutions: We have $$\begin{aligned} \int_{e}^{e^{4}} \frac{1}{x \sqrt{\ln x}} d x & =\int_{e}^{e^{4}}(\ln (x))^{-1 / 2} x^{-1} d x \\ & =\int_{1}^{4} u^{-1 / 2} d u \quad\left(u=\ln x, d u=x^{-1} d x\right) \\ & =\left[2 u^{1 / 2}\right]_{1}^{4} \\ & =2(4)^{1 / 2}-2(1)^{1 / 2} \\ & =2. \end{aligned}$$

10. Over the time interval $1 \leq t \leq 100$ the temperature of a freezer compartment is given by $$f(t)=\frac{t-4}{t^{2}},$$ where $t$ is measured in hours and $f(t)$ is measured in degrees Celsius. What’s the maximum temperature of the freezer over this time interval?

    Solution: Because $f$ is continuous on $[1,100]$ (it’s a rational function whose domain includes the entire interval $[1,100]$), the Extreme-Value Theorem tell us that $f$ attains a maximum value at some number, say $c$, in $[1,100]$. This number $c$ must either be an endpoint of $[1,100]$ or a critical number of $f$ in $(1,100)$. We have $$f^{\prime}(t)=\frac{t^{2}(1)-(t-4) 2 t}{t^{4}}=\frac{8-t}{t^{3}},$$ and thus $8$ is the only critical number of $f$ inside $(1,100)$. The largest of the three numbers $f(1), f(8), f(100)$ will be the maximum value of $f$ on $[1,100]$. Since $f(1)=-3, f(8)=1 / 16$, and $f(100)=96 / 10000< 100 / 10000=1 / 100$, we see the maximum temperature of the freezer over the time interval $[1,100]$ is $1 / 16$ degrees Celsius.