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AP Calculus AB : Derivative at a point

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Example Question #51 : Derivative At A Point

Functionplot

$\begin{align*}&\text{Conclude if the derivative of the function above at the point }\\&x=2\\&\text{is positive, negative, or zero.}\end{align*}$

Possible Answers:

$\text{The derivative is zero.}$

$\text{The derivative is negative.}$

$\text{The derivative is positive.}$

Correct answer:

$\text{The derivative is positive.}$

Explanation:

$\begin{align*}&\text{The derivative of a function at a given point is synonymous}\\&\text{with the slope of a function at a given point. Therefore if}\\&\text{the function increases in value as x increases across a point,}\\&\text{the slope and derivative are positive. Conversely, if the function}\\&\text{decreases in value as x increases across a point, both slope}\\&\text{and derivative are negative. At points where the function is}\\&\text{flat, the slope and derivative are zero.}\\&\text{Studying the slope at }x=2\text{, the function appears to be increasing.}\\&\text{The derivative is positive.}\end{align*}$

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Example Question #151 : Ap Calculus Ab

Functionplot

$\begin{align*}&\text{Conclude if the derivative of the function above at the point }\\&x=-0.2\\&\text{is positive, negative, or zero.}\end{align*}$

Possible Answers:

$\text{The derivative is zero}$

$\text{The derivative is positive.}$

$\text{The derivative is negative.}$

Correct answer:

$\text{The derivative is zero}$

Explanation:

$\begin{align*}&\text{The derivative of a function at a given point is synonymous}\\&\text{with the slope of a function at a given point. Therefore if}\\&\text{the function increases in value as x increases across a point,}\\&\text{the slope and derivative are positive. Conversely, if the function}\\&\text{decreases in value as x increases across a point, both slope}\\&\text{and derivative are negative. At points where the function is}\\&\text{flat, the slope and derivative are zero.}\\&\text{Studying the slope at }x=-0.2\text{, the function appears to be flat.}\\&\text{The derivative is zero}\end{align*}$

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Example Question #53 : Derivative At A Point

Functionplot

$\begin{align*}&\text{Determine whether or not the derivative of the function above at the point }\\&x=-0.8\\&\text{is positive, negative, or zero.}\end{align*}$

Possible Answers:

$\text{The derivative is positive.}$

$\text{The derivative is zero.}$

$\text{The derivative is negative.}$

Correct answer:

$\text{The derivative is positive.}$

Explanation:

$\begin{align*}&\text{The derivative of a function at a given point is synonymous}\\&\text{with the slope of a function at a given point. Therefore if}\\&\text{the function increases in value as x increases across a point,}\\&\text{the slope and derivative are positive. Conversely, if the function}\\&\text{decreases in value as x increases across a point, both slope}\\&\text{and derivative are negative. At points where the function is}\\&\text{flat, the slope and derivative are zero.}\\&\text{Studying the slope at }x=-0.8\text{, the function appears to be increasing.}\\&\text{The derivative is positive.}\end{align*}$

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Example Question #31 : Slope Of A Curve At A Point

Find the slope of the tangent line to the following function at x=3:

f(x)=10x^3+1

Possible Answers:

180

270

181

271

Correct answer:

270

Explanation:

The slope of the tangent line to a curve is given by the first derivative evaluated at the point of interest.

The first derivative of the function is equal to

f'(x)=30x^2

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

To find the slope of the tangent line at x=3, we simply evaluate the first derivative function at x=3:

f'(3)=30(3^2)=270

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Example Question #51 : Derivative At A Point

Find the slope of the curve y=3x^5+9x+3 at the point x=1

Possible Answers:

Correct answer:

Explanation:

To find the slope of the curve at any point, we take its derivative

y'=15x^4+9

We then evaluate ate the point x=1 ,

y'(1)=15(1)^4+9=24

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Example Question #52 : Derivative At A Point

Find k'(1) if $k(x)=\frac{3x^2+4x-2}{10}$ .

Possible Answers:

Correct answer:

Explanation:

First, find the derivative. You should get $k'(x)=\frac{1}{10}(6x+4)$ .

Next, plug in x=1.

You should get k'(1)=1 .

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Example Question #53 : Derivative At A Point

Find the slope of the curve at the specified point:

y=2x^3+4x-10 at the point x=1

Possible Answers:

Correct answer:

Explanation:

To find the slope of the curve, we take the derivative of the function:

$y'=m=\frac{\mathrm{d} }{\mathrm{d} x}(2x^3+4x-10)=6x^2+4$

Evaluating the slope at the specified point, we get

y'(1)=6(1^2)+4=10

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Example Question #58 : Derivative At A Point

Find the normal line to the function $f(x)=\frac{4}{x}$ at the point (1,4) .

Possible Answers:

$y=\frac{x}{4}+3.75$

$y=-\frac{x}{4}+3.75$

$y=-\frac{x}{4}-3.75$

y=x+3.75

$y=\frac{x}{4}-3.75$

Correct answer:

$y=\frac{x}{4}+3.75$

Explanation:

The normal line to a function at a point is the line perpendicular to its tangent line. The normal line to a function has a slope which is the negative reciprocal of its derivative.

For our function $f(x)=\frac{4}{x}$ , the derivative is $f'(x)=\frac{-4}{x^2}$ .

We now may plug in our point of x=1 , $f'(1)=\frac{-4}{1}=-4$ , However, since we seek the negative reciprocal it now becomes $m=\frac{1}{4}$

Our normal line will have the form:

$y-4=\frac{1}{4}(x-1)$

$y=\frac{x}{4}+3.75$

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Example Question #59 : Derivative At A Point

If y=3x^2+2x+2 is the equation of a line, what is the slope the line when x=3

Possible Answers:

Correct answer:

Explanation:

First, we take the derivative of the line, which gives us the slope of the curve at any point

$y'=\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2x+2)=6x+2$

Then, we evaluate the slope at the point x=3 to obtain an actual value

y'(3)=20

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Example Question #54 : Derivative At A Point

Given the equation of the line

y=2x^3+2x-3 , what is slope of the curve when x=1

Possible Answers:

m=6

m=7

m=8

m=12

Correct answer:

m=8

Explanation:

To solve the problem, we simply take the derivative of the function and evaluate it at the specified point. The derivative of a line is the slope of the curve, so we do the following

y'=6x^2+2

y'(1)=m=6(1)^2+2=8

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AP Calculus AB : Derivative at a point

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #51 : Derivative At A Point

Example Question #151 : Ap Calculus Ab

Example Question #53 : Derivative At A Point

Example Question #31 : Slope Of A Curve At A Point

Example Question #51 : Derivative At A Point

Example Question #52 : Derivative At A Point

Example Question #53 : Derivative At A Point

Example Question #58 : Derivative At A Point

Example Question #59 : Derivative At A Point

Example Question #54 : Derivative At A Point

All AP Calculus AB Resources

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