-
The graph of \(\displaystyle y = \frac{x - 3}{x^2 + 4x - 21}\) has
- a vertical asymptote at \(\displaystyle x = -7\) and a removable discontinuity at \(\displaystyle x = 3\).
- a vertical asymptote at \(\displaystyle x = 3\) and a removable discontinuity at \(\displaystyle x = -7\).
- removable discontinuities at both \(\displaystyle x = -7\) and \(\displaystyle x = 3\).
- vertical asymptotes at both \(\displaystyle x = -7\) and \(\displaystyle x = 3\).
- neither removable discontinuities nor vertical asymptotes.
Solution: Factor the denominator: \(x^2+4x-21=(x+7)(x-3)\). Then \(\dfrac{x-3}{(x+7)(x-3)}=\dfrac{1}{x+7}\) for \(x\neq3\). So \(x=3\) is a removable discontinuity and \(x=-7\) is a vertical asymptote.
-
Let \(\displaystyle f(x) = 2x^3 - x^2 + 1\). The tangent line to the graph of \(\displaystyle f(x)\) at \(\displaystyle x = 1\) is parallel to which of the following lines?
- \(\displaystyle y = 5x - 1\)
- \(\displaystyle y = 4x + 3\)
- \(\displaystyle y = 2x + 2\)
- \(\displaystyle y = 4\)
- None of those
Solution: Compute \(f'(x)=6x^2-2x\). At \(x=1\), the slope is \(f'(1)=6-2=4\). Therefore the tangent line is parallel to any line of slope \(4\), namely \(\boxed{y=4x+3}\).
-
The function \(\displaystyle f\) is defined on \(\displaystyle (-2, 3)\). Its graph is given below:
(graph of \(\displaystyle f\), with domain from \(\displaystyle -2\) to \(\displaystyle 3\), piecewise linear and nonlinear segments, with a corner at some point, and open/closed endpoints)
Below are four graphs. One of them is the graph of \(\displaystyle f'\), and one of them is the graph of \(\displaystyle f''\).
(graph labeled (1), piecewise with vertical lines and open/closed endpoints)
(graph labeled (2), piecewise nonlinear)
(graph labeled (3), piecewise, including intervals with constant values and a jump)
(graph labeled (4), piecewise with vertical lines and open/closed endpoints)
Which is the graph of \(\displaystyle f'\), and which is the graph of \(\displaystyle f''\)?
In the choices below, the first number corresponds to the graph of \(\displaystyle f'\), the second one to that of \(\displaystyle f''\).
- 1, 3
- 3, 1
- 1, 4
- 2, 3
- 3, 2
IMAGE
Solution: To identify \(f'\), first look for slope information in the graph of \(f\). On \((0,3)\), the graph of \(f\) is a straight line with slope \(-1\). That means \(f'(x)\) must be the constant value \(-1\) on that interval. So among the candidate graphs, the graph of \(f'\) must be the one that has a horizontal piece at height \(-1\) over \((0,3)\).
Next, on \((-2,0)\), the graph of \(f\) is increasing and curved, so \(f'(x)\) should be positive there and should vary with \(x\), not remain constant. Once we identify the graph that matches both of those features, the graph of \(f''\) is obtained by differentiating that \(f'\)-graph: \(f''\) should record where \(f'\) is increasing, decreasing, or constant.
Matching those qualitative features with the four candidates leads to the correct pair.
-
Given \(\displaystyle f(x) = \frac{\ln(x^3 + x + 1)}{x \arcsin(x)}\), what is \(\displaystyle f'(x)\)?
A
\[\displaystyle \frac{\ln(3x^2 + 1)x \arcsin(x) - \ln(x^3 + x + 1)x \arccos(x)}{(x \arcsin(x))^2}\]
B
\[\displaystyle \frac{\ln(x^3 + x + 1)x \arccos(x) - \ln(3x^2 + 1)x \arcsin(x)}{(x \arcsin(x))^2}\]
C
\[\displaystyle \frac{\ln(x^3 + x + 1)(\arcsin(x) + x \arccos(x)) - \ln(3x^2 + 1)x \arcsin(x)}{(x \arcsin(x))^2}\]
D
\[\displaystyle \frac{\left( \frac{3x^2+1}{x^3+x+1} \right)x \arcsin(x) - \ln(x^3 + x + 1) \left( \arcsin(x) + x \cdot \frac{1}{\sqrt{1-x^2}} \right)}{(x \arcsin(x))^2}\]
E
\[\displaystyle \frac{\ln(x^3 + x + 1)\left( \arcsin(x) + x \cdot \frac{1}{\sqrt{1-x^2}} \right) - \left( \frac{3x^2+1}{x^3+x+1} \right)x \arcsin(x)}{(x \arcsin(x))^2}\]
Solution: Use the quotient rule. Let \(u=\ln(x^3+x+1)\) and \(v=x\arcsin x\). Then \(u'=\dfrac{3x^2+1}{x^3+x+1}\) and \(v'=\arcsin x + x\dfrac{1}{\sqrt{1-x^2}}\). Hence \(f'(x)=\dfrac{u'v-uv'}{v^2}\), which matches choice \(\boxed{D}\).
-
What is the value of
\[\displaystyle \int_{0}^{2} \left(1 + \sqrt{4 - x^2}\right) dx\]
(Hint: Use the following graph of \(\displaystyle y = 1 + \sqrt{4 - x^2}\), which is defined over \(\displaystyle [-2, 2]\).)
(Graph of \(\displaystyle y = 1 + \sqrt{4 - x^2}\))
- \(\displaystyle \pi\)
- \(\displaystyle 2\pi\)
- \(\displaystyle 4\pi\)
- \(\displaystyle \pi + 2\)
- \(\displaystyle 2\pi + 4\)
IMAGE
Solution: On \([0,2]\), the graph \(y=1+\sqrt{4-x^2}\) is a rectangle of area \(2\) plus a quarter-circle of radius \(2\), whose area is \(\pi\). Therefore the integral equals \(\boxed{\pi+2}\).
-
What is \(\displaystyle g'(e)\), if \(\displaystyle g(x) = \int_{1}^{x^2} (t \ln t) dt\) ?
- 0
- 1
- \(\displaystyle e^2\)
- \(\displaystyle 2e^3\)
- \(\displaystyle 4e^3\)
Solution: By the Fundamental Theorem of Calculus and the chain rule, \(g'(x)=\bigl(x^2\ln(x^2)\bigr)\cdot 2x=2x^3\ln(x^2)\). At \(x=e\), this becomes \(2e^3\ln(e^2)=2e^3\cdot2=\boxed{4e^3}\).
-
Compute \(\displaystyle \int \left( x^3 - 8x + x^{2/3} \right) dx\).
- \(\displaystyle \frac{1}{4} x^4 - 4x^2 + \frac{3}{5} x^{5/3} + C\)
- \(\displaystyle 3x^2 - 8 + \frac{2}{3}x^{-1/3} + C\)
- \(\displaystyle x^3 - 8x + x^{2/3} + C\)
- \(\displaystyle 3x^2 - 4x^2 + x^{2/3} + C\)
- \(\displaystyle \frac{1}{4}x^4 - 8x + \frac{3}{5}x^{5/3} + C\)
Solution: Integrate term by term: \(\int x^3dx=\tfrac14x^4\), \(\int -8x\,dx=-4x^2\), and \(\int x^{2/3}dx=\tfrac35x^{5/3}\). Hence an antiderivative is \(\boxed{\tfrac14x^4-4x^2+\tfrac35x^{5/3}+C}\).
-
Using the properties of the definite integral find the value of
\[\displaystyle \int_{3}^{7} (1 - 5f(x)) dx\]
if it is known that
\[\displaystyle \int_{3}^{8} f(x) dx = 10 \text{and} \int_{7}^{8} f(x) dx = 8.\]
- \(\displaystyle -46\)
- \(\displaystyle -10\)
- \(\displaystyle -9\)
- \(\displaystyle -6\)
- \(\displaystyle 2\)
Solution: First find \(\int_3^7 f(x)\,dx=\int_3^8 f(x)\,dx-\int_7^8 f(x)\,dx=10-8=2\). Then \(\int_3^7(1-5f(x))dx=\int_3^7 1\,dx-5\int_3^7 f(x)\,dx=4-10=\boxed{-6}\).
-
Find the following limits. If they do not exist, choose DNE.
- () \(\displaystyle \lim_{x \to 2} \frac{4x^2 - 16}{x - 2}\)
- () \(\displaystyle \lim_{x \to +\infty} x - \sqrt{x^2 + x}\)?
Solution:
- Factor: \(\dfrac{4x^2-16}{x-2}=\dfrac{4(x-2)(x+2)}{x-2}=4(x+2)\) for \(x\neq2\). So the limit is \(\boxed{16}\).
- Rationalize: \(x-\sqrt{x^2+x}=\dfrac{x^2-(x^2+x)}{x+\sqrt{x^2+x}}=\dfrac{-x}{x+\sqrt{x^2+x}}\). Dividing by \(x\) gives the limit \(\boxed{-\dfrac12}\).
-
Given the values of \(\displaystyle f(x)\) and \(\displaystyle f'(x)\) in the table below, and given that
\[\displaystyle h(x) = f(3 + f(x)),\]
find \(\displaystyle h'(1)\).
| x |
f(x) |
f'(x) |
| 1 |
1 |
2 |
| 2 |
4 |
1 |
| 3 |
3 |
5 |
| 4 |
2 |
1/2 |
Solution: Since
\[
h(x)=f(3+f(x)),
\]
this is a composition of functions, so we use the chain rule:
\[
h'(x)=f'(3+f(x))\cdot f'(x).
\]
Now evaluate at \(x=1\). From the table,
\[
f(1)=1,
\]
so
\[
3+f(1)=4.
\]
Also from the table,
\[
f'(4)=\frac12,\qquad f'(1)=2.
\]
Therefore
\[
h'(1)=f'(4)\cdot f'(1)=\frac12\cdot2=1.
\]
Hence
\[
\boxed{h'(1)=1}.
\]
-
- () Find the linear approximation of \(\displaystyle h(x) = \sqrt{x}\) at \(\displaystyle x = 9\).
- () Use the above to estimate \(\displaystyle \sqrt{9.1}\).
Solution:
- The linearization of a function \(h\) at \(x=a\) is
\[
L(x)=h(a)+h'(a)(x-a).
\]
Here \(h(x)=\sqrt{x}\) and \(a=9\). We compute
\[
h(9)=3,\qquad h'(x)=\frac{1}{2\sqrt{x}},\qquad h'(9)=\frac16.
\]
Therefore
\[
L(x)=3+\frac{x-9}{6}.
\]
- To estimate \(\sqrt{9.1}\), substitute \(x=9.1\) into the linearization:
\[
\sqrt{9.1}\approx L(9.1)=3+\frac{9.1-9}{6}=3+\frac{0.1}{6}=3.016\overline6.
\]
So the estimate is
\[
\boxed{3.016\overline6}.
\]
-
Find the slope of the tangent to the curve implicitly defined by the equation
\[\displaystyle y^4 - x y^2 + x^4 = 1\]
at the point (1, 1).
Solution: Differentiate implicitly: \(4y^3y'-(y^2+2xyy')+4x^3=0\). Grouping the \(y'\)-terms gives \((4y^3-2xy)y'=y^2-4x^3\). At \((1,1)\), this yields \((4-2)y'=1-4=-3\), so \(y'=\boxed{-\dfrac32}\).
-
Sketched below is the graph of the function \(\displaystyle y = f(x)\).
[Graph of the function \(\displaystyle y = f(x)\) is shown.]
- () On the graph, sketch the tangent line to \(\displaystyle f(x)\) at \(\displaystyle x = 2\).
- () Use the graph to estimate the value of \(\displaystyle f'(2)\). Show the work that leads to your estimate.
IMAGE
Solution: To estimate \(f'(2)\), we look at the tangent line to the graph at \(x=2\). The tangent line appears to be decreasing, so the derivative should be negative.
A reasonable way to estimate the slope is to choose two convenient points on the tangent line rather than on the original curve. From the picture, moving one unit to the right along the tangent line corresponds to a drop of about two vertical units. That gives the slope estimate
\[
\frac{\Delta y}{\Delta x}\approx\frac{-2}{1}=-2.
\]
So a reasonable estimate is
\[
\boxed{f'(2)\approx -2}.
\]
-
The derivative of the function \(\displaystyle g(x)\) is given below. Both \(\displaystyle g\) and \(\displaystyle g'\) are defined everywhere except at 0 and 1.
\[\displaystyle g'(x) = \frac{x^2 - 4}{x^2 - x}\]
Answer the following questions.
- () What are the \(\displaystyle x\) coordinates of all local minima of \(\displaystyle g(x)\)?
– If there aren't any, write NONE.
ANSWER:________________________________________
- () What are the \(\displaystyle x\) coordinates of all local maxima of \(\displaystyle g(x)\)?
– If there aren't any, write NONE.
ANSWER:________________________________________
Solution: Since \(g'(x)=\dfrac{(x-2)(x+2)}{x(x-1)}\), we study its sign on the intervals cut by \(-2,0,1,2\). The sign pattern is \(+,-,+,-,+\). Hence \(g\) has a local maximum at \(x=-2\) and a local minimum at \(x=2\). The points \(x=0\) and \(x=1\) are not in the domain, so they are not extrema of \(g\).
-
The graph of \(\displaystyle y = f'(x)\), the derivative function of \(\displaystyle f(x)\), is shown below. Assume that \(\displaystyle f(x)\) is defined and continuous on \(\displaystyle [-5, 6]\). Give a complete answer to the following questions.
\[\displaystyle \begin{array}{c} \text{Graph showing } y = f'(x) \text{ with labeled axes and key points marked.} \\ \text{Attention: This is NOT the graph of } f(x). \end{array}\]
- () What are the intervals where \(\displaystyle f(x)\) is concave up?
– If there aren’t any, write NONE.
ANSWER:__________________________________
- () How many inflection points does the graph of \(\displaystyle f(x)\) have?
– If there aren’t any, write 0.
ANSWER:__________________________________
- () What are the intervals where \(\displaystyle f(x)\) is increasing?
– If there aren’t any, write NONE.
ANSWER:__________________________________
IMAGE
Solution: Remember that the picture is the graph of \(f'(x)\), not of \(f(x)\).
\(f(x)\) is concave up where \(f'(x)\) is increasing. From the graph, \(f'\) increases on
\[
\boxed{(-5,-3)\cup(4,6)}.
\]
Inflection points of \(f\) occur where the concavity changes, that is, where \(f'\) changes from increasing to decreasing or from decreasing to increasing. From the graph, this happens twice, so \(f\) has
\[
\boxed{2 \text{ inflection points}}.
\]
Finally, \(f(x)\) is increasing where \(f'(x)>0\), meaning where the graph of \(f'\) lies above the x-axis. From the picture, that occurs on
\[
\boxed{(-4,0)\cup(2,4)\cup(4,6)}.
\]
If one prefers, the last two intervals can be combined as \((2,6)\) while noting that \(f'(4)=0\) at the isolated point \(x=4\).
-
Below is the sign chart of a function \(\displaystyle f\) whose domain is \(\displaystyle [-3,2) \cup (2, \infty)\). Sketch the function as well as possible given the available data.
| x |
3 |
(-3, -1) |
1 |
(-1, 0) |
0 |
(0, 1) |
1 |
(1, 2) |
2 |
(2, +\infty) |
+\infty |
| f |
0 |
|
|
|
0 |
|
|
|
\lim\limits_{x\to2^-} f(x) = +\infty |
|
\lim\limits_{x\to+\infty} f(x) = 1 |
| f' |
+ |
+ |
0 |
|
|
|
0 |
+ |
|
|
|
| f'' |
|
|
|
|
0 |
+ |
+ |
+ |
|
+ |
|
(A coordinate grid for sketching a function is shown.)
IMAGE
Solution: A correct sketch must satisfy the sign chart: zeros at \(x=-3\) and \(x=0\), horizontal tangents at \(x=-1\) and \(x=1\), concave up on the intervals indicated, a vertical asymptote at \(x=2\) with \(f(x)\to+\infty\) as \(x\to2^-\), and a horizontal asymptote \(y=1\) as \(x\to+\infty\). Any graph obeying all of these features earns full credit.
-
What are the absolute maxima and the absolute minima of the function
\[\displaystyle f(x) = x^3 - 3x^2 + 1\]
on the interval [0, 5]?
Solution: Compute \(f'(x)=3x^2-6x=3x(x-2)\). On \([0,5]\), check the endpoints and the critical point \(x=2\). We get \(f(0)=1\), \(f(2)=8-12+1=-3\), and \(f(5)=125-75+1=51\). Therefore the absolute minimum is \(\boxed{-3}\) at \(x=2\), and the absolute maximum is \(\boxed{51}\) at \(x=5\).
-
Compute \(\displaystyle \int x^2 \cos (x^3 + 3)\ dx\).
Solution: Use the substitution \(u=x^3+3\), so \(du=3x^2dx\). Then \(\int x^2\cos(x^3+3)\,dx=\dfrac13\int \cos u\,du=\boxed{\dfrac13\sin(x^3+3)+C}\).
-
Consider the region \(\displaystyle R\) bounded by \(\displaystyle y = x^2 + 1\) and \(\displaystyle y = x\) between \(\displaystyle x = 0\) and \(\displaystyle x = 1\).
- () Sketch \(\displaystyle R\). Make sure to label the graphs.
(Graph with labeled axes and grid, showing space to sketch the described region.)
- () Find its area.
IMAGE
Solution:
- A correct sketch shows \(y=x^2+1\) above \(y=x\) on \([0,1]\).
- The area is \(\int_0^1\bigl((x^2+1)-x\bigr)dx=\left[\dfrac{x^3}{3}+x-\dfrac{x^2}{2}\right]_0^1=\boxed{\dfrac56}\).
-
On a windy day, Camila launches a sunny yellow kite 300 ft into the sky. The wind tugs it horizontally at 25 ft/sec. When the string reaches 500 ft, how fast should she let it out?
Solution: Let \(x\) be the horizontal distance and \(s\) the string length. The kite stays 300 ft high, and the wind pulls it horizontally at \(dx/dt=25\) ft/s. We have \(s^2=x^2+300^2\). When \(s=500\), \(x=400\). Differentiating gives \(2s\,ds/dt=2x\,dx/dt\), so \(ds/dt=\dfrac{x\,dx/dt}{s}=\dfrac{400\cdot25}{500}=\boxed{20\text{ ft/s}}\).
-
With final exams wrapping up, many people will be traveling over the winter break. In preparation for this, people will need to make sure to check the dimensions of their checked and carry-on luggage to make sure that they are not too big. Passengers of many airlines are only allowed to carry a piece of luggage into an airplane if the total of its length, width, and height does not exceed 45 in.
- () Suppose that you wish to carry on a rectangular suitcase whose length is exactly 1.5 times its width, and whose dimensions add up to 45 in. Let \(\displaystyle w\) be the width of the suitcase. Give a formula, \(\displaystyle V(w)\), for the volume (in in\(\displaystyle ^3\)) of such a suitcase in terms of \(\displaystyle w\).
- () Find a reasonable domain for \(\displaystyle w\).
- () Find the value of \(\displaystyle w\) at which the volume of the suitcase \(\displaystyle V(w)\) is maximized.
Solution:
- If the width is \(w\), then the length is \(1.5w\) and the height is \(45-2.5w\). Hence \(V(w)=1.5w^2(45-2.5w)\).
- A reasonable domain is \(0
- Differentiate: \(V'(w)=135w-11.25w^2=11.25w(12-w)\). The interior critical point is \(w=12\), which gives the maximum volume. Thus the suitcase volume is maximized when \(\boxed{w=12\text{ in}}\).