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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #166 : Derivative Review

What is the slope of $f(x)=-x^{2}+5x-2$ at (2,-5) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have

$f(x)=-x^{2}+5x-2$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$

to determine that

$f'(x)=(2)(-1)x^{2-1}+(1)5x^{1-1}-(0)2=-2x+5$ .

We also have a point (2,-5) with a -coordinate , so the slope

f'(2)=-2(2)+5=-4+5=1 .

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Example Question #162 : Derivatives

What is the slope of $f(x)=4x^{2}-3x+2$ at (1,-1) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have

$f(x)=4x^{2}-3x+2$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to determine that

$f'(x)=(2)4x^{2-1}-(1)3x^{1-1}+(0)2=8x-3$ .

We also have a point (1,-1) with a -coordinate , so the slope

f'(1)=8(1)-3=8-3=5 .

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Example Question #168 : Derivative Review

What is the slope of $f(x)=-x^{2}+11x+4$ at (4,-5) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have

$f(x)=-x^{2}+11x+4$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$

to determine that

$f'(x)=(2)(-1)x^{2-1}+(1)11x^{1-1}+(0)4=-2x+11$ .

We also have a point (4,-5) with a -coordinate , so the slope

f'(4)=-2(4)+11=-8+11=3 .

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Example Question #163 : Derivative Review

Find f'(1) for

$f(x)=x^3+x^2+x+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}$

Possible Answers:

$\infty$

Correct answer:

Explanation:

In order to find f'(1) , we first find f'(x) .

$f(x)=x^3+x^2+x+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}$

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

Now we plug in 1 to get

$f'(1)=3(1)^2+2(1)+1-\frac{1}{(1)^2}-\frac{2}{(1)^3}-\frac{3}{(1)^4}$

f'(1)=3+2+1-1-2-3=0

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Example Question #163 : Derivatives

What is the slope of $f(x)=3x^{2}-14$ at the point (4,-2) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have $f(x)=3x^{2}-14$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$

to determine that

$f'(x)=(2)3x^{2-1}-(0)14=6x$ .

We also have a point (4,-2) with a -coordinate , so the slope

f'(4)=6(4)=24 .

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Example Question #171 : Derivative Review

What is the slope of $f(x)=2x^{2}+11x-14$ at the point (-1,6) ?

Possible Answers:

None of the above

$-\frac{1}{7}$

$\frac{1}{7}$

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have $f(x)=2x^{2}+11x-14$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$

to determine that

$f(x)=(2)2x^{2-1}+(1)11x^{1-1}-(0)14=4x+11$ .

We also have a point (-1,6) with a -coordinate , so the slope

f(-1)=4(-1)+11=-4+11=7 .

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Example Question #172 : Derivative Review

What is the slope of $f(x)=\frac{1}{3}x^{2}-4x$ at the point (7,2) ?

Possible Answers:

$\frac{3}{2}$

$-\frac{2}{3}$

$-\frac{3}{2}$

$\frac{2}{3}$

None of the above

Correct answer:

$\frac{2}{3}$

Explanation:

We define slope as the first derivative of a given function.

Since we have $f(x)=\frac{1}{3}x^{2}-4x$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$

to determine that

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

We also have a point (7,2) with a -coordinate , so the slope

$f'(7)=\frac{2}{3}(7)-4=\frac{14}{3}-\frac{12}{3}=\frac{2}{3}$ .

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Example Question #1291 : Calculus Ii

Calculate the derivative of $f(x)=\frac{x^3}{3}-2x^2+x-3$ at the point x=3 .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

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

Calculate f'(x)

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

f'(x)=x^2 - 4x + 1

Then, plug in the value of x and evaluate

f'(x=3)=3^2-4*3+1=-2

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Example Question #1292 : Calculus Ii

Calculate the derivative of $f(x)=\frac{x^3}{12}-4x^2+9x-13$ at the point x=3 .

Possible Answers:

-12.75

-11.75

-13.75

-14.75

Correct answer:

-12.75

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

$f(x)=\frac{x^3}{12}-4x^2+9x-13$

Calculate f'(x)

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

$f'(x)=\frac{3x^2}{12}-8x+9$

Then, plug in the value of x and evaluate

$f'(x=3)=\frac{3^2}{4}-8*3+9= \frac{9}{4}-24+9=-12.75$

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Example Question #174 : Derivative Review

Calculate the derivative of $f(x)=\frac{9x^3}{16}+3x^2+2$ at the point x=4 .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x)

Then, replace the value of x with the given point and evaluate

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

$f(x)=\frac{9x^3}{16}+3x^2+2$

Calculate f'(x)

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

$f'(x)=\frac{9*3*x^2}{16}+3*2*x$

Then, plug in the value of x and evaluate

$f'(x=4)=\frac{27*4^2}{16}+6*4=\frac{27*16}{16}+24=27+24=51$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #166 : Derivative Review

Example Question #162 : Derivatives

Example Question #168 : Derivative Review

Example Question #163 : Derivative Review

Example Question #163 : Derivatives

Example Question #171 : Derivative Review

Example Question #172 : Derivative Review

Example Question #1291 : Calculus Ii

Example Question #1292 : Calculus Ii

Example Question #174 : Derivative Review

All Calculus 2 Resources

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