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AP Calculus BC : AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #11 : Limits

Limitplot

$\begin{align*}&\text{Which of the values most likely represents the value of f(x) at}\\&x=\infty\end{align*}$

Possible Answers:

$\text{Nonexistent.}$

$-\infty$

$-\frac{15}{17}$

Correct answer:

$-\infty$

Explanation:

$\begin{align*}&\text{Although actually observing a function at an infinite value}\\&\text{of x is impossible, we can often determine via observation if}\\&\text{the function approaches a certain limit. Looking at the values}\\&\text{the function approaches as the absolute value of x increases}\\&\text{gives a clue. Things to look out for are if the function continuously}\\&\text{climbs/decreases, flattens, or has unchecked oscillatory behavior.}\\&\text{If the function continuously climbs/decreases, it likely has}\\&\text{a limit that approaches positive/negative infinity. If it flattens,}\\&\text{the limit is a real number. If it oscillates without control,}\\&\text{the limit does not exist.}\\&\text{The limit of }f(\infty)\text{ is most likely }-\infty\end{align*}$

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Example Question #1 : Limits And Continuity

Rational_graph

The graph above is a sketch of the function . For what intervals is continuous?

Possible Answers:

$(-\infty, -2) \cup \(2, \infty)$

$(-\infty, -2) \cup (-2, 2)$

$(-\infty, \infty)$

$(-\infty, -2) \cup (-2, 2) \cup \(2, \infty)$

$(-2, 2) \cup \(2, \infty)$

Correct answer:

$(-\infty, -2) \cup (-2, 2) \cup \(2, \infty)$

Explanation:

For a function to be continuous at a point , f(a) must exist and $\lim_{x \rightarrow a}f(x)=f(a)$ .

This is true for all values of except x=-2 and x=2 .

Therefore, the interval of continuity is $(-\infty, -2) \cup (-2, 2) \cup \(2, \infty)$ .

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Example Question #1 : Limits And Continuity

Consider the piecewise function:

$y=\left\{\begin{matrix} 1;\: x<-5\\0;\:x>-5 \end{matrix}\right.$

What is $\lim_{x \to -5}$ ?

Possible Answers:

$\frac{1}{2}$

0<x<1

Limit does not exist.

Correct answer:

Limit does not exist.

Explanation:

The piecewise function

$\left\{\begin{matrix} 1;\: x<-5\\0;\:x>-5 \end{matrix}\right.$

indicates that is one when is less than five, and is zero if the variable is greater than five. At x=-5 , there is a hole at the end of the split.

The limit does not indicate whether we want to find the limit from the left or right, which means that it is necessary to check the limit from the left and right. From the left to right, the limit approaches 1 as approaches negative five. From the right, the limit approaches zero as approaches negative five.

Since the limits do not coincide, the limit does not exist for $\lim_{x \to -5}$ .

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Example Question #3 : Limits And Continuity

Consider the function $f(x)=\frac{x^2-7x+12}{x-3}$ .

Which of the following statements are true about this function?

I. $f(x)\ is\ continuous\ at\ x = 3.$

II. $\lim_{x\to3}f(x)=-1$

III. $\lim_{x\to3}f(x)\ does\ not\ exist$

Possible Answers:

I and III

II only

I and II

III only

Correct answer:

I and II

Explanation:

For a function to be continuous at a particular point, the limit of the function at that point must be equal to the value of the function at that point.

First, notice that

$f(3)=\frac{3^2-7(3)+12}{3-3}=\frac{(x-3)(x-4)}{(x-3)}=(x-4)=(3-4)=-1$ .

This means that the function is continuous everywhere.

Next, we must compute the limit. Factor and simplify f(x) to help with the calculation of the limit.

$f(x)=\frac{x^2-7x+12}{x-3}=\frac{(x-4)(x-3)}{(x-3)}=x-4$

$\lim_{x\to3}(x-4)=-1$

Thus, the limit as x approaches three exists and is equal to , so I and II are true statements.

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Example Question #1 : Asymptotic And Unbounded Behavior

A cylinder of height 4in and radius 1in is expanding. The radius increases at a rate of $0.1\frac{in}{s}$ and its height increases at a rate of $0.2\frac{in}{s}$ . What is the rate of growth of its surface area?

Possible Answers:

$0.4\pi\frac{in^2}{s}$

$2.8\pi\frac{in^2}{s}$

$1.4\pi\frac{in^2}{s}$

$1.6\pi\frac{in^2}{s}$

$0.2\pi\frac{in^2}{s}$

Correct answer:

$1.6\pi\frac{in^2}{s}$

Explanation:

The surface area of a cylinder is given by the formula:

$A=2\pi r^2+2\pi rh$

To find the rate of growth over time, take the derivative of each side with respect to time:

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt} + 2\pi r \frac{dh}{dt}$

Therefore, the rate of growth of surface area is:

$\frac{dA}{dt}=4\pi (1in)\left(0.1\frac{in}{s}\right)+2\pi(1in)(4in)\left(0.1\frac{in}{s}\right)+2\pi(1in)\left(0.2\frac{in}{s}\right)$

$\frac{dA}{dt}=(0.4\pi+0.8\pi+0.4\pi)\frac{in^2}{s}=1.6\pi\frac{in^2}{s}$

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Example Question #2 : Asymptotic And Unbounded Behavior

The rate of growth of the population of Reindeer in Norway is proportional to the population. The population increased from 9876 to 10381 between 2013 and 2015. What is the expected population in 2030?

Possible Answers:

42581

38914

15150

49362

53880

Correct answer:

15150

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased from 9876 to 10381 between 2013 and 2015, we can solve for this constant of proportionality:

$10381=9876e^{k(2015-2013)}$

$\frac{10381}{9876}=e^{2k}$

$2k=ln(\frac{10381}{9876})$

$k=\frac{ln(\frac{10381}{9876})}{2}=0.0252$

Using this, we can calculate the expected value from 2015 to 2030:

$P=10381e^{(2030-2015)(0.0252)} \approx 15150$

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Example Question #3 : Asymptotic And Unbounded Behavior

The rate of decrease due to poaching of the elephants in unprotected Sahara is proportional to the population. The population in one region decreased from 1038 to 817 between 2010 and 2015. What is the expected population in 2017?

Possible Answers:

493

779

619

742

527

Correct answer:

742

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population decreased from 1038 to 817 between 2010 and 2015, we can solve for this constant of proportionality:

$817=1038e^{k(2015-2010)}$

$\frac{817}{1038}=e^{5k}$

$5k=ln(\frac{817}{1038})$

$k=\frac{ln(\frac{817}{1038})}{5}=-0.048$

Using this, we can calculate the expected value from 2015 to 2017:

$P=817e^{(2017-2015)(-0.048)} \approx 742$

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Example Question #1 : Identifying Asymptotes Graphically

Asymptoteplot

$\begin{align*}&\text{Decide which of the following functions, most likely represents the graph above.}\end{align*}$

Possible Answers:

$\frac{(13x - 9)}{(17x - 2)}$

$\frac{(13x + 9)}{(17x + 2)}$

$\frac{(17x - 2)}{(13x - 9)}$

$\frac{(17x + 2)}{(13x + 9)}$

Correct answer:

$\frac{(13x - 9)}{(17x - 2)}$

Explanation:

$\begin{align*}&\text{Observation of the graph shows that there’s a sharp increase}\\&\text{and decrease on either side of a particular x-value. This type}\\&\text{of behavior is observed when there’s a vertical asymptote. An}\\&\text{asymptote is a value that a function may approach, but will}\\&\text{never actually attain. In the case of vertical asymptotes, this}\\&\text{behavior occurs if the function approaches infinity for a given}\\&\text{x-value, often when a zero value appears in a denominator. Noting}\\&\text{this rule, the above function has a zero in the denominator}\\&\text{at a definite point:}\\&x=\frac{2}{17}\\&\text{We find a zero denominator for the function: }\\&\frac{(13x - 9)}{(17x - 2)}\end{align*}$

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Example Question #1 : Asymptotic And Unbounded Behavior

Asymptoteplot

$\begin{align*}&\text{Which of the four following functions is depicted in the graph above?}\end{align*}$

Possible Answers:

$\frac{(14x + 19)}{(4x - 18)}$

$\frac{(4x - 18)}{(14x + 19)}$

$\frac{(14x - 19)}{(4x + 18)}$

$\frac{(4x + 18)}{(14x - 19)}$

Correct answer:

$\frac{(4x + 18)}{(14x - 19)}$

Explanation:

$\begin{align*}&\text{Observation of the graph shows that there’s a sharp increase}\\&\text{and decrease on either side of a particular x-value. This type}\\&\text{of behavior is observed when there’s a vertical asymptote. An}\\&\text{asymptote is a value that a function may approach, but will}\\&\text{never actually attain. In the case of vertical asymptotes, this}\\&\text{behavior occurs if the function approaches infinity for a given}\\&\text{x-value, often when a zero value appears in a denominator. Noting}\\&\text{this rule, the above function has a zero in the denominator}\\&\text{at a definite point centered around:}\\&x=1.3\\&\text{We find a zero denominator for the function: }\\&\frac{(4x + 18)}{(14x - 19)}\end{align*}$

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Example Question #1 : Asymptotic And Unbounded Behavior

Asymptoteplot

$\begin{align*}&\text{Which of the four following functions is depicted in the graph above?}\end{align*}$

Possible Answers:

$-\frac{(5x + 12)}{(6x - 10)}$

$-\frac{(6x - 10)}{(5x + 12)}$

$\frac{(12x - 5)}{(10x + 6)}$

$\frac{(18x + 6)}{(12x - 5)}$

Correct answer:

$-\frac{(5x + 12)}{(6x - 10)}$

Explanation:

$\begin{align*}&\text{Notice that as x increases, the function graphed appears to}\\&\text{level out, flattening towards a certain value. This value is}\\&\text{what is known as a horizontal asymptote. An asymptote is a value}\\&\text{that a function approaches, but never actually reaches. Think}\\&\text{of a horizontal asymptote as the limit of a function as x approaches}\\&\text{infinity. In such a case, as x approaches infinity, any constants}\\&\text{added or subtracted in the numerator and denominator become}\\&\text{irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=-0.8\\&\text{This matches the ratio of coefficients for the function: }\\&-\frac{(5x + 12)}{(6x - 10)}\end{align*}$

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Limits

Example Question #1 : Limits And Continuity

Example Question #1 : Limits And Continuity

Example Question #3 : Limits And Continuity

Example Question #1 : Asymptotic And Unbounded Behavior

Example Question #2 : Asymptotic And Unbounded Behavior

Example Question #3 : Asymptotic And Unbounded Behavior

Example Question #1 : Identifying Asymptotes Graphically

Example Question #1 : Asymptotic And Unbounded Behavior

Example Question #1 : Asymptotic And Unbounded Behavior

All AP Calculus BC Resources

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