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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 #64 : Functions, Graphs, And Limits

Asymptoteplot

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

Possible Answers:

$-\frac{(3x - 2)}{(10x - 11)}$

$\frac{(10x - 11)}{(3x - 2)}$

$-\frac{(10x - 11)}{(3x - 2)}$

$\frac{(3x - 2)}{(10x - 11)}$

Correct answer:

$\frac{(3x - 2)}{(10x - 11)}$

Explanation:

$\begin{align*}&\text{Notice that as x increases, the function graphed appears to}\\&\text{level out, flattening towards a certain value. This value is}\\&\text{what is known as a horizontal asymptote. An asymptote is a value}\\&\text{that a function approaches, but never actually reaches. Think}\\&\text{of a horizontal asymptote as the limit of a function as x approaches}\\&\text{infinity. In such a case, as x approaches infinity, any constants}\\&\text{added or subtracted in the numerator and denominator become}\\&\text{irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=0.3\\&\text{This matches the ratio of coefficients for the function: }\\&\frac{(3x - 2)}{(10x - 11)}\end{align*}$

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Example Question #5 : Asymptotic And Unbounded Behavior

Asymptoteplot

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

Possible Answers:

$-\frac{(19x + 9)}{(17x - 5)}$

$-\frac{(17x - 5)}{(19x + 9)}$

$\frac{(17x - 5)}{(19x + 9)}$

$\frac{(19x + 9)}{(17x - 5)}$

Correct answer:

$-\frac{(19x + 9)}{(17x - 5)}$

Explanation:

$\begin{align*}&\text{Notice that as x increases, the function graphed appears to}\\&\text{level out, flattening towards a certain value. This value is}\\&\text{what is known as a horizontal asymptote. An asymptote is a value}\\&\text{that a function approaches, but never actually reaches. Think}\\&\text{of a horizontal asymptote as the limit of a function as x approaches}\\&\text{infinity. In such a case, as x approaches infinity, any constants}\\&\text{added or subtracted in the numerator and denominator become}\\&\text{irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=-1.12\\&\text{This matches the ratio of coefficients for the function: }\\&-\frac{(19x + 9)}{(17x - 5)}\end{align*}$

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Example Question #1 : Identifying Asymptotes Graphically

Asymptoteplot

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

Possible Answers:

$-\frac{(15x - 4)}{(x + 13)}$

$-\frac{(15x + 4)}{(x - 13)}$

$-\frac{(x + 13)}{(15x - 4)}$

$-\frac{(x - 13)}{(15x + 4)}$

Correct answer:

$-\frac{(x - 13)}{(15x + 4)}$

Explanation:

$\begin{align*}&\text{Observation of the graph shows that there’s a sharp increase}\\&\text{and decrease on either side of a particular x-value. This type}\\&\text{of behavior is observed when there’s a vertical asymptote. An}\\&\text{asymptote is a value that a function may approach, but will}\\&\text{never actually attain. In the case of vertical asymptotes, this}\\&\text{behavior occurs if the function approaches infinity for a given}\\&\text{x-value, often when a zero value appears in a denominator. Noting}\\&\text{this rule, the above function has a zero in the denominator}\\&\text{at a definite point centered approximately around:}\\&x=-0.3\\&\text{We find a zero denominator for the function: }\\&-\frac{(x - 13)}{(15x + 4)}\end{align*}$

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Example Question #1 : Identifying Asymptotes Graphically

Asymptoteplot

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

Possible Answers:

$\frac{(15x - 1)}{(9x - 11)}$

$\frac{(9x + 11)}{(15x + 1)}$

$\frac{(9x - 11)}{(15x - 1)}$

$\frac{(15x + 1)}{(9x + 11)}$

Correct answer:

$\frac{(9x - 11)}{(15x - 1)}$

Explanation:

$\begin{align*}&\text{Observation of the graph shows that there’s a sharp increase}\\&\text{and decrease on either side of a particular x-value. This type}\\&\text{of behavior is observed when there’s a vertical asymptote. An}\\&\text{asymptote is a value that a function may approach, but will}\\&\text{never actually attain. In the case of vertical asymptotes, this}\\&\text{behavior occurs if the function approaches infinity for a given}\\&\text{x-value, often when a zero value appears in a denominator. Noting}\\&\text{this rule, the above function has a zero in the denominator}\\&\text{at a definite point centered approximately around:}\\&x=0.1\\&\text{We find a zero denominator for the function: }\\&\frac{(9x - 11)}{(15x - 1)}\end{align*}$

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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:

$-\frac{(12x - 20)}{(17x)}$

$\frac{(17x)}{(12x - 20)}$

$-\frac{(17x)}{(12x - 20)}$

$\frac{(12x - 20)}{(17x)}$

Correct answer:

$\frac{(12x - 20)}{(17x)}$

Explanation:

$\begin{align*}&\text{Notice that as x increases, the function graphed appears to}\\&\text{level out, flattening towards a certain value. This value is}\\&\text{what is known as a horizontal asymptote. An asymptote is a value}\\&\text{that a function approaches, but never actually reaches. Think}\\&\text{of a horizontal asymptote as the limit of a function as x approaches}\\&\text{infinity. In such a case, as x approaches infinity, any constants}\\&\text{added or subtracted in the numerator and denominator become}\\&\text{irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=0.71\\&\text{This matches the ratio of coefficients for the function: }\\&\frac{(12x - 20)}{(17x)}\end{align*}$

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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:

$-\frac{(16x - 13)}{(9x - 4)}$

$\frac{(16x - 13)}{(9x - 4)}$

$\frac{(9x - 4)}{(16x - 13)}$

$-\frac{(9x - 4)}{(16x - 13)}$

Correct answer:

$\frac{(16x - 13)}{(9x - 4)}$

Explanation:

$\begin{align*}&\text{Notice that as x increases, the function graphed appears to}\\&\text{level out, flattening towards a certain value. This value is}\\&\text{what is known as a horizontal asymptote. An asymptote is a value}\\&\text{that a function approaches, but never actually reaches. Think}\\&\text{of a horizontal asymptote as the limit of a function as x approaches}\\&\text{infinity. In such a case, as x approaches infinity, any constants}\\&\text{added or subtracted in the numerator and denominator become}\\&\text{irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=1.78\\&\text{This matches the ratio of coefficients for the function: }\\&\frac{(16x - 13)}{(9x - 4)}\end{align*}$

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Example Question #1 : Identifying Asymptotes Graphically

Asymptoteplot

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

Possible Answers:

$-\frac{(15x - 16)}{(3x + 14)}$

$-\frac{(3x + 14)}{(15x - 16)}$

$\frac{(15x - 16)}{(3x + 14)}$

$\frac{(3x + 14)}{(15x - 16)}$

Correct answer:

$\frac{(3x + 14)}{(15x - 16)}$

Explanation:

$\begin{align*}&\text{Notice that as x increases, the function graphed appears to}\\&\text{level out, flattening towards a certain value. This value is}\\&\text{what is known as a horizontal asymptote. An asymptote is a value}\\&\text{that a function approaches, but never actually reaches. Think}\\&\text{of a horizontal asymptote as the limit of a function as x approaches}\\&\text{infinity. In such a case, as x approaches infinity, any constants}\\&\text{added or subtracted in the numerator and denominator become}\\&\text{irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=0.2\\&\text{This matches the ratio of coefficients for the function: }\\&\frac{(3x + 14)}{(15x - 16)}\end{align*}$

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Example Question #71 : Functions, Graphs, And Limits

Asymptoteplot

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

Possible Answers:

$\frac{(20x + 10)}{(12x - 10)}$

$-\frac{(20x + 10)}{(12x - 10)}$

$-\frac{(12x - 10)}{(20x + 10)}$

$\frac{(12x - 10)}{(20x + 10)}$

Correct answer:

$-\frac{(20x + 10)}{(12x - 10)}$

Explanation:

$\begin{align*}&\text{Notice that as x increases, the function graphed appears to}\\&\text{level out, flattening towards a certain value. This value is}\\&\text{what is known as a horizontal asymptote. An asymptote is a value}\\&\text{that a function approaches, but never actually reaches. Think}\\&\text{of a horizontal asymptote as the limit of a function as x approaches}\\&\text{infinity. In such a case, as x approaches infinity, any constants}\\&\text{added or subtracted in the numerator and denominator become}\\&\text{irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=-1.67\\&\text{This matches the ratio of coefficients for the function: }\\&-\frac{(20x + 10)}{(12x - 10)}\end{align*}$

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Example Question #72 : Functions, Graphs, And Limits

Asymptoteplot

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

Possible Answers:

$-\frac{(16x - 9)}{(3x + 18)}$

$-\frac{(16x + 9)}{(6x - 18)}$

$-\frac{(6x - 18)}{(16x + 9)}$

$-\frac{(3x + 18)}{(16x - 9)}$

Correct answer:

$-\frac{(6x - 18)}{(16x + 9)}$

Explanation:

$\begin{align*}&\text{Notice that as x increases, the function graphed appears to}\\&\text{level out, flattening towards a certain value. This value is}\\&\text{what is known as a horizontal asymptote. An asymptote is a value}\\&\text{that a function approaches, but never actually reaches. Think}\\&\text{of a horizontal asymptote as the limit of a function as x approaches}\\&\text{infinity. In such a case, as x approaches infinity, any constants}\\&\text{added or subtracted in the numerator and denominator become}\\&\text{irrelevant. What matters is the power of x in the denominator}\\&\text{and the numerator; if those are the same, then the coefficients}\\&\text{define the asymptote. We see that the function flattens towards:}\\&f(\infty)=-\frac{3}{8}\\&\text{This matches the ratio of coefficients for the function: }\\&-\frac{(6x - 18)}{(16x + 9)}\end{align*}$

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Example Question #1 : Derivatives

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Possible Answers:

{ ${}{x:x<e^{\frac{-3}{2}}}$ }

{ ${}{x:x>e^{\frac{-3}{2}}}$ }

{ ${}{x:0<x<e^{\frac{-3}{2}}}$ }

f(x) is never concave upward.

$\left \{ x:x>\ln\frac{3}{2}\right \}$

Correct answer:

{ ${}{x:0<x<e^{\frac{-3}{2}}}$ }

Explanation:

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

$=-2x\ln(x)+-x^2(\frac{1}{x})$

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

$=-(2\ln x +1)+-x\cdot\frac{2}{x}=-2\ln x-1-2=-2\ln x-3$

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

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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 #64 : Functions, Graphs, And Limits

Example Question #5 : Asymptotic And Unbounded Behavior

Example Question #1 : Identifying Asymptotes Graphically

Example Question #1 : Identifying Asymptotes Graphically

Example Question #1 : Identifying Asymptotes Graphically

Example Question #1 : Identifying Asymptotes Graphically

Example Question #1 : Identifying Asymptotes Graphically

Example Question #71 : Functions, Graphs, And Limits

Example Question #72 : Functions, Graphs, And Limits

Example Question #1 : Derivatives

All AP Calculus BC Resources

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