Question
1. $\lim _{x \rightarrow \infty} \frac{20 x^2-13 x+5}{5-4 x^3}$ is
(A) $-5$
(B) 0
(C) 1
(D) $\infty$
Answer/Explanation
Ans:B
Use the Rational Function Theorem
A rational function is of the form
$
f(x)=\frac{P(x)}{Q(x)}
$
where $P(x)$ and $Q(x)$ are polynomials. The domain of $f$ is the set of all reals for which $Q(x) \neq 0$.
Question
2. $\lim _{h \rightarrow 0} \frac{\ln (2+h)-\ln 2}{h}$ is
(A) $\ln 2$
(B) $\frac{1}{2}$
(C) $\frac{1}{\ln 2}$
(D) $\infty$
Answer/Explanation
Ans:B
Note that $\lim _{h \rightarrow 0} \frac{\ln (2+h)-\ln 2}{h}=f^{\prime}(2)$, where $f(x)=\ln x$.
Question
3. If $y=e^{-x^2}$, then $y^{\prime \prime}(0)$ equals
(A) 2
(B) 1
(C) 0
(D) $-2$
Answer/Explanation
Ans:D
Since $y^{\prime}=-2 x e^{-x^2}$, therefore $y^{\prime \prime}=-2\left(x \cdot e^{-x^2} \cdot(-2 x)+e^{-x^2}\right)$. Replace $x$ by 0 .
Question
Questions 4 and 5. Use the following table, which shows the values of the differentiable functions $f$ and $g$.
4. The average rate of change of function $f$ on $[1,4]$ is
(A) $7 / 6$
(B) $4 / 3$
(C) $15 / 8$
(D) $15 / 4$
Answer/Explanation
Ans:B
$\frac{f(4)-f(1)}{4-1}=\frac{6-2}{4-1}=\frac{4}{3}$
Question
5. If $h(x)=g(f(x))$ then $h^{\prime}(3)=$
(A) $1 / 2$
(B) 1
(C) 4
(D) 6
Answer/Explanation
Ans:B
$h^{\prime}(3)=g^{\prime}(f(3)) \cdot f^{\prime}(3)=g^{\prime}(4) \cdot f^{\prime}(3)=\frac{1}{2} \cdot 2$
Question
6. The derivative of a function $f$ is given for all $x$ by
$
f^{\prime}(x)=x^2(x+1)^3(x-4)^2
$
The set of $x$-values for which $f$ is a relative minimum is
(A) $\{0,-1,4\}$
(B) $\{-1\}$
(C) $\{0,4\}$
(D) $\{0,-1\}$
Answer/Explanation
Ans:B
Since $f(x)$ exists for all $x$, it must equal 0 for any $x_0$ for which $f$ is a relative minimum, and it must also change sign from negative to positive as $x$ increases through $x_0$. For the given derivative, this only occurs at $x=-1$.
Question
7. If $y=\frac{x-3}{2-5 x}$, then $\frac{d y}{d x}$ equals
(A) $\frac{17-10 x}{(2-5 x)^2}$
(B) $\frac{13}{(2-5 x)^2}$
(C) $\frac{-13}{(2-5 x)^2}$
(D) $\frac{17}{(2-5 x)^2}$
Answer/Explanation
Ans:C
By the Quotient Rule (formula (6) on page 115),
$
\frac{d y}{d x}=\frac{(2-5 x)(1)-(x-3)(-5)}{(2-5 x)^2}
$
The Quotient Rule. $\lim _{x \rightarrow c}\left(\frac{f(x)}{g(x)}\right)=\frac{\lim _{x \rightarrow c} f(x)}{\lim _{x \rightarrow c} g(x)}=\frac{B}{D}$, provided that $D \neq 0$
Question
8. The maximum value of the function $f(x)=x e^{-x}$ is
(A) $\frac{1}{e}$
(B) 1
(C) $-1$
(D) $-e$
Answer/Explanation
Ans:A
Here, $f(x)$ is $e^{-x}(1-x)$; $f$ has maximum value when $x=1$.
Question
9. Which equation has the slope field shown below?
(A) $\frac{d y}{d x}=\frac{5}{y}$
(B) $\frac{d y}{d x}=\frac{5}{x}$
(C) $\frac{d y}{d x}=\frac{x}{y}$
(D) $\frac{d y}{d x}=5 y$
Answer/Explanation
Ans:A
Note that (1) on a horizontal line the slope segments are all parallel, so the slopes there are all the same and $\frac{d y}{d x}$ must depend only on $y ;(2)$ along the $x$-axis (where $y=0$ ) the slopes are infinite; and (3) as $y$ increases, the slope decreases.
Question
Questions 10-12. The graph below shows the velocity of an object moving along a line, for $0 \leq t \leq 9$.
10. At what time does the object attain its maximum acceleration?
(A) $2<t<5$
(B) $t=6$
(C) $t=8$
(D) $8<t<9$
Answer/Explanation
Ans:D
Acceleration is the derivative (the slope) of velocity $v$; $v$ is largest on $8<t<9$.
Question
The object is farthest from the starting point at $t=$
(A) 5
(B) 6
(C) 8
(D) 9
Answer/Explanation
Ans:B
Velocity $v$ is the derivative of position; because $v>0$ until $t=6$ and $v<0$ thereafter, the position increases until $t=6$ and then decreases; since the area bounded by the curve above the axis is larger than the area below the axis, the object is farthest from its starting point at $t=6$.
Question
At $t=8$, the object was at position $x=10$. At $t=5$, the object’s position was $x=$
(A) 5
(B) 7
(C) 13
(D) 15
Answer/Explanation
Ans:C
From $t=5$ to $t=8$, the displacement (the integral of velocity) can be found by evaluating definite integrals based on the areas of two triangles: $\frac{1}{2}(1)(2)-\frac{1}{2}(2)(4)=-3$. Thus, if $K$ is the object’s position at $t=5$, then $K-3=10$ at $t=8$
Question
$\int_0^2 \frac{x^2-3 x+7}{x+3} d x=$
(A) $\frac{4}{3}$
(B) $-10+25 \ln \frac{5}{3}$
(C) $2+7 \ln \frac{5}{3}$
(D) $\frac{32}{5} \ln 5$
Answer/Explanation
Ans:B
(B) When we have a rational function in the integrand and the degree of the numerator is greater than the degree of the denominator, we need to rewrite the integrand using long polynomial division.
$
\begin{gathered}
\frac{-\left(x^2+3 x\right)}{-6 x+7} \\
\frac{-(-6 x-18)}{25} \\
\int_0^2\left(x-6+\frac{25}{x+3}\right) d x=\left.\left(\frac{x^2}{2}-6 x+25 \ln |x+3|\right)\right|_0 ^2 \\
=(2-12+25 \ln 5)-25 \ln 3=-10+25 \ln \frac{5}{3}
\end{gathered}
$
Question
14. You are given two twice-differentiable functions, $f(x)$ and $g(x)$. The table above gives values for $f(x)$ and $g(x)$ and their first and second derivatives at $x=1$. Find $\lim _{x \rightarrow 1} \frac{2 f(x)-6 g(x)}{4 x^2-4 e^{3(x-1)}}$.
(A) $-3$
(B) 1
(C) 2
(D) nonexistent
Answer/Explanation
Ans:A
The functions are twice differentiable, meaning the functions and their first derivatives are continuous, so we can use substitution when finding limits:
$\lim _{x \rightarrow 1}(2 f(x)-6 g(x))=2(1)-6\left(\frac{1}{3}\right)=0 \quad \lim _{x \rightarrow 1}\left(4 x^2-4 e^{3(x-1)}\right)=4(1)^2-4 e^{3(1-1)}=0$
Since the limit of both the numerator and the denominator are zero, we can use L’Hospital’s Rule:
$
\lim _{x \rightarrow 1} \frac{2 f(x)-6 g(x)}{4 x^2-4 e^{3(x-1)}}=\lim _{x \rightarrow 1} \frac{2 f^{\prime}(x)-6 g^{\prime}(x)}{8 x-12 e^{3(x-1)}}=\frac{2(0)-6(-2)}{8(1)-12 e^{3(1-1)}}=\frac{12}{-4}=-3
$
Question
15. A differentiable function has the values shown in this table:
Estimate $f(2.1)$.
(A) $0.34$
(B) $1.56$
(C) $1.70$
(D) $1.91$
Answer/Explanation
Ans:C
$f^{\prime}(2.1) \approx \frac{f(2.2)-f(2.0)}{2.2-2.0}$
Question
16. If $A=\int_0^1 e^{-x} d x$ is approximated using various sums with the same number of subdivisions, and if $L, R$, and $T$ denote, respectively, left. Riemann Sum, right Riemann Sum, and trapezoidal sum, then it follows that
(A) $R \leq A \leq T \leq L$
(B) $R \leq T \leq A \leq L$
(C) $L \leq T \leq A \leq R$
(D) $L \leq A \leq T \leq R$
Answer/Explanation
Ans:A
(A) $f(x)=e^{-x}$ is decreasing and concave upward.
Question
17. The number of vertical tangents to the graph of $y^2=x-x^3$ is
(A) 3
(B) 2
(C) 1
(D) 0
Answer/Explanation
Ans:A
Implicit differentiation yields $2 y y^{\prime}=1-3 x^2$; so $\frac{d y}{d x}=\frac{1-3 x^2}{2 y}$. At a vertical tangent, $\frac{d y}{d x}$ is undefined; $y$ must therefore equal 0 and the numerator must be nonzero. The original equation with $y=0$ is $0=x-x^3$, which has three solutions.
Question
$\int_0^6 f(x-1) d x=$
(A) $\int_1^7 f(x) d x$
(B) $\int_{-1}^5 f(x) d x$
(C) $\int_{-1}^5 f(x+1) d x$
(D) $\int_1^7 f(x+1) d x$
Answer/Explanation
Ans:B
(B) Let $t=x-1$; then $t=-1$ when $x=0, t=5$ when $x=6$, and $d t=d x$.
Question
The equation of the curve shown below is $y=\frac{4}{1+x^2}$. What does the area of the shaded region equal?
(A) $8-\Pi$
(B) $8-2 \pi$
(C) $8-4 \Pi$
(D) $8-4 \ln 2$
Answer/Explanation
Ans:B
The required area, $A$, is given by the integral
$
2 \int_0^1\left(4-\frac{4}{1+x^2}\right) d x=\left.2\left(4 x-4 \tan ^{-1} x\right)\right|_0 ^1=2\left(4-4 \cdot \frac{\pi}{4}\right)
$
Question
Over the interval $0 \leq x \leq 10$, the average value of the function $f$ shown below
(A) is $6.10$
(B) is $6.25$
(C) does not exist, because $f$ is not continuous
(D) does not exist, because $f$ is not integrable
Answer/Explanation
Ans:A
The average value is $\frac{1}{10-0} \int_0^{10} f(x) d x$. The definite integral represents the sum of the areas of a trapezoid and a rectangle: $\frac{1}{2}(8+3)(6)=4(7)=61$.
Question
If $f(x)=2 f(x)$ and $f(2)=1$, then $f(x)=$
(A) $e^{2 x-4}$
(B) $e^{2 x}+1-e^4$
(C) $e^{4-2 x}$
(D) $e^{x^2-4}$
Answer/Explanation
Ans:A
Solve the differential equation $\frac{d y}{d x}=2 y$ by separation of variables: $\frac{d y}{y}=2 d x$ yields $y=c e^{2 x}$. The initial condition yields $1=c e^{2 \cdot 2}$; so $c=e^{-4}$ and $y=e^{2 x-4}$.
Question
If $f(t)=\int_0^{t^2} \frac{1}{1+x^2} d x$, then $f(t)$ equals
(A) $\frac{1}{1+t^2}$
(B) $\frac{2 t}{1+t^2}$
(C) $\frac{1}{1+t^4}$
(D) $\frac{2 t}{1+t^4}$
Answer/Explanation
Ans:D
$\frac{d}{d u} \int_0^u \frac{1}{1+x^2} d x=\frac{1}{1+u^2}$. When $u=t^2$,
$
\frac{d}{d t} \int_0^u \frac{1}{1+x^2} d x=\frac{1}{1+u^2} \frac{d u}{d t}=\frac{1}{1+t^4}(2 t)
$
Question
23. The curve $x^3+x \tan y=27$ passes through $(3,0)$. Use the tangent line there to estimate the value of $y$ at $x=3.1$. The value is
(A) $-2.7$
(B) $-0.9$
(C) 0
(D) $0.1$
Answer/Explanation
Ans:B
By implicit differentiation, $3 x^2+x \sec ^2 y \frac{d y}{d x}+\tan y=0$. At $(3,0), \frac{d y}{d x}=-9$; so an equation of the tangent line at $(3,0)$ is $y=-9(x-3)$.
Question
$\int(\sqrt{x}-2) x^2 d x=$
(A) $\frac{2}{3} x^{3 / 2}-2 x+C$
(B) $\frac{5}{2} x^{3 / 2}-4 x+C$
(C) $\frac{2}{5} x^{5 / 2}-\frac{2}{3} x^3+C$
(D) $\frac{2}{7} x^{7 / 2}-\frac{2}{3} x^3+C$
Answer/Explanation
Ans:D
$\int(\sqrt{x}-2) x^2 d x=\int\left(x^{5 / 2}-2 x^2\right) d x=\frac{2}{7} x^{7 / 2}-\frac{2}{3} x^3+C$
Question
25. The graph of a function $y=f(x)$ is shown above. Which is true?
(A) $\lim _{x \rightarrow 1} f(x)=-\infty$
(B) $\lim _{x \rightarrow-\infty} f(x)=\pm 1$
(C) $\lim _{x \rightarrow-2} f(x)=0$
(D) $\lim _{x \rightarrow \infty} f(x)=0$
Answer/Explanation
Ans:D
The graph shown has the $x$-axis as a horizontal asymptote.
Question
A function $f(x)$ equals $\frac{x^2-x}{x-1}$ for all $x$ except $x=1$. For the function to be continuous at $x=1$, the value of $f(1)$ must be
(A) 0
(B) 1
(C) 2
(D) $\infty$
Answer/Explanation
Ans:B
Since $\lim _{x \rightarrow 1} f(x)=1$, to render $f(x)$ continuous at $x=1, f(1)$ must be defined to be 1 .
Question
$\int_0^1 \frac{e^x}{\left(3-e^x\right) 2} d x$ equals
(A) $-2 \ln (3-e)$
(B) $\frac{1-e}{2(3-e)}$
(C) $\frac{1}{2(3-e)}$
(D) $\frac{e-1}{2(3-e)}$
Answer/Explanation
Ans:D
$-\int_0^1\left(3-e^x\right)^{-2}\left(-e^x d x\right)=\left.\frac{1}{3-e^x}\right|_0 ^1=\frac{1}{3-e}-\frac{1}{2}$
Question
Suppose $f(x)=\int_0^x \frac{4+t}{t^2+4} d t$. It follows that
(A) $f$ increases for all $x$
(B) $f$ has a critical point at $x=0$
(C) $f$ has a local min at $x=-4$
(D) $f$ has a local max at $x=-4$
Answer/Explanation
Ans:C
Note that $f^{\prime}(x)=\frac{4+x}{x^2+4}$, so $f$ has a critical value at $x=-4$. As $x$ passes through $-4$, the sign of $f$ changes from – to $+$, so $f$ has a local minimum at $x=-4$.
Question
The graph of $f(x)$ consists of two line segments as shown above. If $g(x)=f^{-1}(x)$, the inverse function of $f(x)$, find $g$ (4).
(A) $\frac{1}{5}$
(B) $\frac{2}{3}$
(C) $\frac{3}{2}$
(D) 5
Answer/Explanation
Ans:C
(C) Given the point $(a, b)$ on function $f(x),\left(f^{-1}\right)^{\prime}(b)=\frac{1}{f^{\prime}(a)}$. Note that the slope of the graph of $f(x)$ at $x=5$ is $\frac{2}{3}$, so $g^{\prime}(4)=\frac{1}{f^{\prime}(5)}=\frac{1}{2 / 3}=\frac{3}{2}$.
Question
30. Choose the integral that is the limit of the Riemann Sum: $\lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\left(\frac{3 k}{n}+2\right)^2 \cdot\left(\frac{3}{n}\right)\right)$.
(A) $\int_2^5(x+2)^2 d x$
(B) $\int_0^3(3 x+2)^2 d x$
(C) $\int_2^5 x^2 d x$
(D) $\int_0^3 x^2 d x$
Answer/Explanation
Ans:C
From the Riemann Sum, we see $\Delta x=\frac{3}{n}$, then $k \cdot \Delta x=\frac{3 k}{n}$. Notice that the term involving $k$ in the Riemann Sum is equal to $\frac{3 k}{n}$. Thus, we try $x_k=\frac{3 k}{n}+2$, so $a=2$ and $\Delta x=\frac{b-2}{n}=\frac{3}{n}$, so $b=5$. Since $x_k$ replaces $x, f(x)=x^2$ giving the integral $\int_2^5 x^2 d x$.
Part B
Question
An object moving along a line has velocity $v(t)=t \cos t-\ln (t+2)$, where $0 \leq t \leq 10$. The object achieves its maximum speed when $t$ is approximately
(A) $5.107$
(B) $6.419$
(C) $7.550$
(D) $9.538$
Answer/Explanation
Ans:D
Use your calculator to graph velocity against time. Speed is the absolute value of velocity. The greatest deviation from $v=0$ is at $t=c$. With a calculator, $c=9.538$.
Question
32. The graph of $f$, which consists of a quarter-circle and two line segments, is shown above. At $x=2$, which of the following statements is true?
(A) $f$ is not continuous.
(B) $f$ is continuous but not differentiable.
(C) $f$ has a local maximum.
(D) The graph of $f$ has a point of inflection.
Answer/Explanation
Ans:D
Because $f$ changes from increasing to decreasing, $f^{\prime \prime}$ changes from positive to negative and thus the graph of $f$ changes concavity.
Question
33. Let $H(x)=\int_0^x f(t) d t$, where $f$ is the function whose graph appears below.
The local linearization of $H(x)$ near $x=3$ is $H(x) \approx$
(A) $-2 x+8$
(B) $2 x-4$
(C) $-2 x+4$
(D) $2 x-8$
Answer/Explanation
Ans:D
$H(3)=\int_0^3 f(t) d t$. We evaluate this definite integral by finding the area of a trapezoid (negative) and a triangle: $H(3)=-\frac{1}{2}(2+1)(2)+\frac{1}{2}(1)(2)=-2$, so the tangent line passes through the point $(3,-2)$. The slope of the line is $H^{\prime}(3)=f(3)=2$, so an equation of the line is $y-(-2)=2(x-3)$.
Question
34. The table shows the speed of an object, in feet per second, at various times during a 12 -second interval.
Estimate the distance the object travels, using the midpoint method with 3 subintervals.
(A) 100 feet
(B) 110 feet
(C) 112 feet
(D) 114 feet
Answer/Explanation
Ans:C
The distance is approximately $14(6)+8(2)+3(4)$.
Question
35. In a marathon, when the winner crosses the finish line many runners are still on the course, some quite far behind. If the density of runners $x$ miles from the finish line is given by $R(x)=20\left[1-\cos \left(1+0.03 x^2\right)\right]$ runners per mile, how many are within 8 miles of the finish line?
(A) 30
(B) 40
(C) 157
(D) 166
Answer/Explanation
Ans:D
$\int_0^8 R(x) d x=166.396$
Question
36. Which best describes the behavior of the function $y=\arctan \left(\frac{1}{\ln x}\right)$ at $x=1$ ?
(A) It has a jump discontinuity.
(B) It has an infinite discontinuity.
(C) It has a removable discontinuity.
(D) It is continuous.
Answer/Explanation
Ans:A
(A) Selecting an answer for this question from your calculator graph is unwise. In some windows the graph may appear continuous; in others there may seem to be cusps, or a vertical asymptote. Put the calculator aside. Find
$
\lim _{x \rightarrow 1^{+}}\left(\arctan \left(\frac{1}{\ln x}\right)\right)=\frac{\pi}{2} \text { and } \lim _{x \rightarrow 1^{-}}\left(\arctan \left(\frac{1}{\ln x}\right)\right)=-\frac{\pi}{2}
$
These limits indicate the presence of a jump discontinuity in the function at $x=1$.
Question
37. Let $G(x)=[f(x)]^2$. In an interval around $x=a$, the graph of $f$ is increasing and concave downward, while $G$ is decreasing. Which describes the graph of $G$ there?
(A) concave downward
(B) concave upward
(C) point of inflection
(D) quadratic
Answer/Explanation
Ans:B
(B) We are given that (1) $f(a)>0$; (2) $f^{\prime}(a)<0$; and (3) $G^{\prime}(a)<0$. Since $G^{\prime}(x)=2 f(x) \cdot f(x)$, therefore $G^{\prime}(a)=2 f(a)$. $f(a)$. Conditions (1) and (3) imply that $(4) f(a)<0$.
Since $G^{\prime \prime}(x)=2\left[f(x) \cdot f^{\prime \prime}(x)+(f(x))^2\right]$, therefore $G^{\prime \prime}(a)=2\left[f(a) f^{\prime \prime}(a)+(f(a))^2\right]$. Then the sign of $G^{\prime \prime}(a)$ is $2[(-) \cdot(-)+$ $(+)]$ or positive, where the minus signs in the parentheses follow from conditions (4) and (2).
Question
38. The value of $c$ for which $f(x)=x+\frac{c}{x}$ has a local minimum at $x=3$ is
(A) $-9$
(B) 0
(C) 6
(D) 9
Answer/Explanation
Ans:D
Since $f^{\prime}(x)=1-\frac{c}{x^2}$, it equals 0 for $x=\pm \sqrt{c}$. When $x=3, c=9$; this yields a minimum since $f^{\prime}(3)>0$.
Question
39. The function $g$ is a differentiable function. It is known that $g^{\prime}(x) \leq 4$ for $3 \leq x \leq 10$ and that $g(7)=8$. Which of the following could be true?
I. $g(5)=0$
II. $g(8)=-4$
III. $g(9)=17$
(A) I only
(B) II only
(C) I and II only
(D) I and III only
Answer/Explanation
Ans:C
We can define function $g$ as $g(x)=g(7)+\int_7^x g^{\prime}(t) d t$. We will use this definition for $g(x)$ and the fact that we can find the maximum value of an integral over an interval by using one rectangle with the height equal to the maximum value of the function over the interval and the width equal to the width of the interval. We were given the maximum value of $g^{\prime}(x)$ on $3 \leq x \leq 10$.
I. $g(5)=g(7)+\int_7^5 g^{\prime}(x) d x \leq 8+(4)(-2)=0$; therefore $g(5)=0$ is possible.
II. $g(8)=g(7)+\int_7^8 g^{\prime}(x) d x \leq 8+(4)(1)=12$; therefore $g(8)=-4$ is possible.
III. $g(9)=g(7)+\int_7^9 g^{\prime}(x) d x \leq 8+(4)(2)=16$; therefore $g(9)=17$ is not possible.
Question
40. At which point on the graph of $y=f(x)$ shown above is $f(x)<0$ and $f^{\prime}(x)>0$ ?
(A) $A$
(B) $B$
(C) $C$
(D) $D$
Answer/Explanation
Ans:A
The curve falls when $f^{\prime}(x)<0$ and is concave up when $f^{\prime \prime}(x)>0$.
Question c
Let $f(x)=x^5+1$, and let $g$ be the inverse function of $f$. What is the value of $g^{\prime}(0)$ ?
(A) $-1$
(B) $\frac{1}{5}$
(C) $-\frac{1}{5}$
(D) $g^{\prime}(0)$ does not exist
Answer/Explanation
Ans:B
$g^{\prime}(y)=\frac{1}{f^{\prime}(x)}=\frac{1}{5 x^4}$. To find $g^{\prime}(0)$, find $x$ such that $f(x)=0$. By inspection, $x=-1$, so
$
g^{\prime}(0)=\frac{1}{5(-1)^4}=\frac{1}{5}
$
Question
The hypotenuse $A B$ of a right triangle $A B C$ is 5 feet, and one leg, $A C$, is decreasing at the rate of 2 feet per second. The rate, in square feet per second, at which the area is changing when $A C=3$ is
(A) $\frac{7}{4}$
(B) $-\frac{3}{2}$
(C) $-\frac{\frac{7}{4}}{4}$
(D) $-\frac{7}{2}$
Answer/Explanation
Ans:C
It is given that $\frac{d x}{d t}=-2$; you want $\frac{d A}{d t}$, where $A=\frac{1}{2} x y$.
$
\frac{d A}{d t}=\frac{1}{2}\left(x \frac{d y}{d t}+y \frac{d x}{d y}\right)=\frac{1}{2}\left[3 \cdot \frac{d y}{d t}+y \cdot(-2)\right]
$
Since $y^2=25-x^2$, it follows that $2 y \frac{d y}{d t}=-2 x \frac{d x}{d t}$ and, when $x=3, y=4$ and $\frac{d y}{d t}=\frac{3}{2}$. Then $\frac{d A}{d t}=-\frac{7}{4}$.
Question
At how many points on the interval $[0, \Pi]$ does $f(x)=2 \sin x+\sin 4 x$ satisfy the Mean Value Theorem?
(A) 1
(B) 2
(C) 3
(D) 4
Answer/Explanation
Ans:D
The function $f(x)=2 \sin x+\sin 4 x$ is graphed above.
Since $f(0)=f(\Pi)$ and $f$ is both continuous and differentiable, Rolle’s Theorem predicts at least one $c$ in the interval such that $f(c)=0$. There are four points in $[0, \Pi]$ of the calculator graph above where the tangent is horizontal.
Question
If the radius $r$ of a sphere is increasing at a constant rate, then the rate of increase of the volume of the sphere is
(A) constant
(B) increasing
(C) decreasing
(D) decreasing for $r<1$ and increasing for $r>1$
Answer/Explanation
Ans:B
Since $\frac{d r}{d t}=k$, a positive constant, $\frac{d V}{d t}=4 \pi r^2 \frac{d r}{d t}=4 \pi r^2 k=c r^2$, where $c$ is a positive constant. Then $\frac{d^2 V}{d t^2}=2 c r \frac{d r}{d t}=2 c r k$, which is also positive.
Question
The rate at which a purification process can remove contaminants from a tank of water is proportional to the amount of contaminant remaining. If $20 \%$ of the contaminant can be removed during the first minute of the process and $98 \%$ must be removed to make the water safe, approximately how long will the decontamination process take?
(A) 2 minutes
(B) 5 minutes
(C) 7 minutes
(D) 18 minutes
Answer/Explanation
Ans:D
If $Q(t)$ is the amount of contaminant in the tank at time $t$ and $Q_0$ is the initial amount, then
$
\frac{d Q}{d t}=k Q \text { and } Q(t)=Q_0 e^{k t}
$
Since $Q(1)=0.8 Q_0, 0.8 Q_0=Q_0 e^{k \cdot 1}, 0.8=e^k$, and
$
Q(t)=Q_0(0.8)^t
$
We seek $t$ when $Q(t)=0.02 Q_0$. Thus,
$
0.02 Q_0=Q_0(0.8)^t
$
and $t \approx 17.53$ minutes
Section
Question
This is the graph of $f′(x)$.
A function $f$ is defined on the interval $[0,4]$ with $f(2)=3, f(x)=e^{\sin x}-2 \cos (3 x)$, and $f^{\prime}(x)=(\cos x) \cdot e^{\sin x}+6$ $\sin (3 x)$
(a) Find all values of $x$ in the interval where $f$ has a critical point. Classify each critical point as a local maximum, local minimum, or neither. Justify your answers.
(b) On what subinterval(s) of $(0,4)$, if any, is the graph of $f$ concave down? Justify your answer.
(c) Write an equation for the line tangent to the graph of $f$ at $x=1.5$.
Answer/Explanation
(a) The graph of $f$ has critical points when $f(x)=0$. This occurs at points $a$ and $b$ on the graph of $f(x)$ above, where $a=0.283$ and $b=3.760$.
At $x=0.283$, the graph of $f$ has a local minimum because $f(x)$ changes from negative to positive at that value.
At $x=3.760$, the graph of $f$ has a local maximum because $f(x)$ changes from positive to negative at that value.
(b) $f^{\prime}(x)=0$ at points $p, q$, and $r$ on the graph of $f(x)$ because that is where the graph of $f(x)$ has horizontal tangent lines, where $p=1.108, q=2.166$, and $r=3.082$. Note, you could also graph $f^{\prime \prime}(x)$ and find the roots of the graph.
To determine the intervals where $f$ is concave down, look for decreasing intervals on $f^{\prime \prime}(x)$ or look for intervals on the graph of $f^{\prime \prime}(x)$ where $f^{\prime \prime}(x)<0$.
The intervals where $f$ is concave down are $(p, q)=(1.108,2.166)$ and $(r, 4)=(3.082,4)$.
(c) $f(1.5)=3.1330726$
$
f(1.5)=f(2)+\int_2^{1.5} f^{\prime}(x) d x=3+\int_2^{1.5} f^{\prime}(x) d x=2.1409738
$
An equation of the tangent line at $x=1.5$ is $y=2.141+3.133(x-1.5)$
Question
2. The rate of sales of a new software product is given by the differentiable and increasing function $S(t)$, where $S$ is measured in units per month and $t$ is measured in months from the initial release date. The software company recorded these sales data:
(a) Use the data in the table to estimate $S^{\prime}(7)$. Show the work that leads to your answer. Indicate units of measure.
(b) Use a left Riemann Sum with the four subintervals indicated in the table to estimate the total number of units of software sold during the first 9 months of sales. Is this an underestimate or an overestimate of the total number of software units sold? Give a reason for your answer.
(c) For $1 \leq t \leq 9$, must there be a time $t$ when the rate of sales is increasing at $100 \frac{\text { units }}{\text { month }}$ per month? Justify your answer.
(d) For $9 \leq t \leq 12$, the rate of sales is modeled by the function $R(t)=120 \cdot(2)^{x / 3}$. Given that the actual sales in the first 9 months is 3636 software units, use this model to find the number of units of the software sold during the first 12 months of sales. Round your answer to the nearest whole unit.
Answer/Explanation
(a) $S^{\prime}(7) \approx \frac{S(8)-S(6)}{8-6}=\frac{763-490}{2}=136.5 \frac{\text { units }}{\text { month }}$ per month.
(b) The total number of units of software sold is given by $\int_1^9 S(t) d t$.
$
\begin{aligned}
\int_1^9 S(t) d t & =S(1) \cdot(3-1)+S(3) \cdot(6-3)+S(6) \cdot(8-6)+S(8) \cdot(9-8) \\
& =154(2)+232(3)+490(2)+763(1)=2747
\end{aligned}
$
This is an underestimate because we used a left Riemann Sum on an increasing function.
(c) Yes, because $\frac{S(9)-S(1)}{9-1}=\frac{954-154}{8}=100$, and $S(t)$ is differentiable and, therefore, continuous, so the Mean Value Theorem guarantees that $S^{\prime}(t)=100 \frac{\text { units }}{\text { month }}$ per month for some value of $t$ on the interval $(1,9)$.
(d) Total Sales $=3636+\int_9^{12} R(t) d t=7790.961718$
The company sold approximately 7791 software units in the first 12 months of sales.
Question
The graph of function $y=f(x)$ passes through the point $(1,1)$ and satisfies the differential equation $\frac{d y}{d x}=\frac{6 x^2-4}{y}$
(a) Sketch the slope field for the differential equation at the 12 indicated points on the axes provided.
(b) Find an equation of the line tangent to $f(x)$ at the point $(1,1)$ and use the linear equation to estimate $f(1.2)$.
(c) Solve the differential equation, and find the particular solution for $y=f(x)$ that passes through the point $(1,1)$.
Answer/Explanation
AB/BC 3. (a) The slope field for the twelve points indicated is:
The slopes are given in the table below. Be sure that the segments you draw are correct relative to the other slopes in the slope field with respect to the steepness of the segments.
(b) At (1,1), $\frac{d y}{d x}=2$, so the tangent line is $y=1+2(x-1)$. Using this linear equation, $f(1.2) \approx 1+2(1.2-1)=$
$1.4$.
(c) The differential equation $\frac{d y}{d x}=\frac{6 x^2-4}{y}$ is separable.
$
\begin{aligned}
\int y d y & =\int\left(6 x^2-4\right) d x \\
\frac{y^2}{2} & =2 x^3-4 x+C_1 \\
y & =\pm \sqrt{4 x^3-8 x+C_2}, \text { where } C_2=2 C_1
\end{aligned}
$
Since $f(x)$ passes through $(1,1)$, it must be true that $1=\pm \sqrt{4(1)^3-8(1)+C_2}$.
Thus $C_2=5$, and the positive square root is used.
The solution is $f(x)=\sqrt{4 x^3-8 x+5}$.
Question
Let $R$ represent the first-quadrant region bounded by the $y$-axis and the curves $y=2^x$ and $y=8 \cos \frac{\pi x}{6}$, as shown in the graph.
(a) Find the area of region $R$.
(b) Set up, but do not evaluate, an integral expression for the volume of the solid formed when $R$ is rotated around the $x$-axis.
(c) Set up, but do not evaluate, an integral expression for the volume of the solid whose base is $R$ and all cross sections in planes perpendicular to the $x$-axis are squares.
Answer/Explanation
AB 4. (a) Draw a vertical element of area, as shown.
$
\begin{aligned}
\Delta A & =\left(y_{\text {top }}-y_{\text {bottom }}\right) \Delta x=\left(8 \cos \frac{\pi x}{6}-2^x\right) \Delta x \\
A & =\int_0^2\left(8 \cos \frac{\pi x}{6}-2^x\right) d x \\
& =\frac{6}{\pi} \cdot 8 \int_0^2 \cos \frac{\pi x}{6} d x-\int_0^2 2^x d x \\
& =\left.\frac{48}{\pi} \cdot \sin \frac{\pi x}{6}\right|_0 ^2-\left.\frac{2^x}{\ln 2}\right|_0 ^2 \\
& =\frac{48}{\pi}\left(\sin \frac{\pi}{3}-\sin 0\right)-\left(\frac{2^2}{\ln 2}-\frac{2^0}{\ln 2}\right) \\
& =\frac{24 \sqrt{3}}{\pi}-\frac{3}{\ln 2}
\end{aligned}
$
(b) Use washers; then
$
\begin{gathered}
\Delta V=\left(r_2^2-r_1^2\right) \Delta x=\pi\left(y_{\text {top }}^2-y_{\text {bottom }}^2\right) \Delta x \\
V=\pi \int_0^2\left[\left(8 \cos \frac{\pi x}{6}\right)^2-\left(2^x\right)^2\right] d x
\end{gathered}
$
(c)
See the figure above.
$
\begin{aligned}
\Delta V & =s^2 \Delta x=\left(y_{\text {top }}-y_{\text {bottom }}\right)^2 \Delta x \\
V & =\int_0^2\left(8 \cos \frac{\pi x}{6}-2^x\right)^2 d x
\end{aligned}
$
Question
5. The graph of the function $f$ is given above. $f$ is twice differentiable and is defined on the interval $-8 \leq x \leq 6$. The function $g$ is also twice differentiable and is defined as $g(x)=\int_{-2}^x f(q) d q$.
(a) Write an equation for the line tangent to the graph of $g(x)$ at $x=-8$. Show the work that leads to your answer.
(b) Using your tangent line from part (a), approximate $g(-7)$. Is your approximation greater than or less than $g(-$ 7)? Give a reason for your answer.
(c) Evaluate: $\int_{-5}^4\left(2 f^{\prime}(x)-3\right) d x$. Show the work that leads to your answer.
(d) Find the absolute minimum value of $g$ on the interval $-8 \leq x \leq 6$. Justify your answer.
Answer/Explanation
(a) $g(-8)=\int_{-2}^{-8} f(q) d q=-\int_{-8}^{-2} f(q) d q=-\left(-\frac{1}{2}(2)(4)+\frac{1}{2}(4)(2)\right)=0 g^{\prime}(x)=f(x)=g^{\prime}(-8)$ $=f(-8)=-4$
Tangent line: $y=0-4(x-(-8)) \Rightarrow y=-4(x+8)$
(b) $g(-7) \approx y(-7)=-4((-7)+8)-4$
$g^{\prime}(x)=f(x)$ is increasing on the interval $[-8,-7]$. Thus, $g^{\prime \prime}(x)>0$ on $[-8,-7]$ and $g(x)$ is concave up on this interval. Therefore, the estimate using the tangent line is less than $g(-7)$ because the tangent line lies below the graph on a concave up portion of the graph.
(c) $\int_{-5}^4\left(2 f^{\prime}(x)-3\right) d x=\left.(2 f(x)-3 x)\right|_{-5} ^4=(2 f(4)-3(4))-(2 f(-5)-3(-5))$
$
=(2(-3)-12)-(2(2)+15)=-37
$
(d) We need to perform the Candidates Test to find the absolute minimum on a closed interval.
First, the candidates are the critical points and the endpoints of the interval.
Critical points: $g^{\prime}(x)=f(x)=0$ when $x=-6,-2,2$, 5. Endpoints: $x=-8,6$
We can eliminate $x=-2,2$ because neither is a local minimum, but we need to check the function value for the remaining candidates.
$g(-8)=0$ (we calculated this in part (a))
$
\begin{aligned}
& g(-6)=\int_{-2}^{-6} f(x) d x=-\int_{-6}^{-2} f(x) d x=-\left(\frac{1}{2}(4)(2)\right)=-4 \\
& g(5)=\int_{-2}^5 f(x) d x=\frac{1}{2} \cdot \pi(2)^2-\frac{1}{2}(3)(3)=2 \pi-\frac{9}{2}>0 \\
& g(6)=\int_{-2}^6 f(x) d x=g(5)+\int_5^6 f(x) d x=\left(2 \pi-\frac{9}{2}\right)+\frac{1}{2}(1)(3)=2 \pi-3>0
\end{aligned}
$
Therefore, the minimum value of $g(x)$ on the interval $[-8,6]$ is $-4$, and it occurs at $x=-6$.
Question
A particle moves along the $x$-axis from $t=0$ to $t=12$. The velocity function of the particle is $v(t)=1+2 \sin \left(\frac{\pi}{6} t\right)$. The initial position of the particle is $x=4$ when $t=0$.
(a) When is the particle moving to the right from $t=0$ to $t=12$ ?
(b) What is the speed of the particle at $t=9$ ?
(c) Find the acceleration function of the particle. Is the speed of the particle increasing or decreasing at time $t=$ 6 ? Explain your reasoning.
(d) Find the position of the particle at time $t=6$.
Answer/Explanation
(а) $v(t)=1+2 \sin \left(\frac{\pi}{6} t\right)=0 \Rightarrow \sin \left(\frac{\pi}{6} t\right)=-\frac{1}{2} \Rightarrow \frac{\pi}{6} t=\frac{7 \pi}{6}, \frac{11 \pi}{6} \Rightarrow t=7,11$
The particle moves to the right when $v(t)>0$; this will occur when $0 \leq t<7$ and $11<t \leq 12$.
(b) The speed of the particle is the absolute value of the velocity. $v(9)=1+2 \sin \left(\frac{9 \pi}{6}\right)=1+2 \sin \left(\frac{3 \pi}{2}\right)=1+2(-1)=-1$. Therefore, the speed at $t=9$ is 1. $a(t)=v^{\prime}(t)=2 \cos \left(\frac{\pi}{6} t\right) \cdot \frac{\pi}{6}=\frac{\pi}{3} \cos \left(\frac{\pi}{6} t\right)$
(c)
$
v(6)=1+2 \sin \left(\frac{6 \pi}{6}\right)=1>0 \quad a(6)=\frac{\pi}{3} \cos \left(\frac{6 \pi}{6}\right)=-\frac{\pi}{3}<0
$
The speed is decreasing at $t=6$ because the velocity and acceleration have opposite signs.
(d) Using the Fundamental Theorem of Calculus:
$
\begin{aligned}
x(6) & =x(0)+\int_0^6 x^{\prime}(t) d t \Rightarrow x(6)=4+\int_0^6 v(t) d t \\
& =4+\int_0^6\left(1+2 \sin \left(\frac{\pi}{6} t\right)\right) d t=4+\left.\left(t-\frac{2 \cos \left(\frac{\pi}{6} t\right)}{\frac{\pi}{6}}\right)\right|_0 ^6 \\
& =4+\left.\left(t-\frac{12}{\pi} \cos \left(\frac{\pi}{6} t\right)\right)\right|_0 ^6=4+\left(\left(6-\frac{12}{\pi} \cos (\pi)\right)-\left(0-\frac{12}{\pi} \cos (0)\right)\right) \\
& =4+\left(\left(6+\frac{12}{\pi}\right)-\left(-\frac{12}{\pi}\right)\right)=10+\frac{24}{\pi}
\end{aligned}
$