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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #482 : Rate

What is the slope at x=0 given the following function: $f(x) = \sin(x^2+2x-3)$

Possible Answers:

2.98

-1.98

3.2

Correct answer:

-1.98

Explanation:

In order to find the slope of a function, you must find its derivative.

In this case, we must find the derivative of the following: $f(x) = \sin(x^2+2x-3)$

That is done by doing the following: $f'(x) = \cos (x^2+2x-3) ((2)x^{(2-1)} + (1)2x^{(1-1)} + 0)$

Therefore, the derivative is: $f'(x) = \cos (x^2+2x-3) (2x + 2)$

Then, you plug x=0 into the derivative: $f'(0) = \cos ((0)^2+2(0)-3) (2(0) + 2)$

Therefore, the answer is: -1.98

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Example Question #483 : Rate

What is the slope at x=0 given the following function: f(x) = ln(2x^3 + e^x)

Possible Answers:

Correct answer:

Explanation:

In order to find the slope of a function, you must find its derivative.

In this case, we must find the derivative of the following: f(x) = ln(2x^3 + e^x)

That is done by doing the following: $f'(x) = \frac{1}{2x^3+e^x} ((3)2x^{(3-1)} + e^x)$

Therefore, the derivative is: $f'(x) = \frac{6x^2+e^x}{2x^3 + e^x}$

Then, you plug x=0 into the derivative: $f'(0) = \frac{6(0)^2+e^(0)}{2(0)^3 + e^(0)}$

Therefore, the answer is:

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Example Question #3311 : Calculus

What is the slope at x=1 given the following function: $f(x) = e^{4x}$

Possible Answers:

345.8

218.4

134.5

234.9

Correct answer:

218.4

Explanation:

In order to find the slope of a function, you must find its derivative.

In this case, we must find the derivative of the following: $f(x) = e^{4x}$

That is done by doing the following: $f'(x) = e^{4x} ((1)4x^{(1-1)})$

Therefore, the derivative is: $f'(x) = 4e^{4x}$

Then, you plug x=1 into the derivative: $f'(1) = 4e^{4(1)}$

Therefore, the answer is: 218.4

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Example Question #401 : How To Find Rate Of Change

What is the slope at x=1 given the following function: $f(x) = \sin (ln(2x))$

Possible Answers:

2.85

0.77

-0.87

0.67

Correct answer:

0.77

Explanation:

In order to find the slope of a function, you must find its derivative.

In this case, we must find the derivative of the following: $f(x) = \sin (ln(2x))$

That is done by doing the following: $f'(x) = \cos (ln(2x)) (\frac{1}{2x}) ((1)2x^{(1-1)})$

Therefore, the derivative is: $f'(x) = \frac{\cos (ln(2x))}{x}$

Then, you plug x=1 into the derivative: $f'(1) = \frac{\cos (ln(2(1)))}{(1)}$

Therefore, the answer is: 0.77

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Example Question #402 : How To Find Rate Of Change

What is the slope at $x=\frac{1}{2}$ given the following function: $f(x) = (6x+3)^3 + \cos(4x)$

Possible Answers:

132.6

320.4

263.7

156.9

-293.8

Correct answer:

320.4

Explanation:

In order to find the slope of a function, you must find its derivative.

In this case, we must find the derivative of the following: $f(x) = (6x+3)^3 + \cos(4x)$

That is done by doing the following: $f'(x) = {\color{Blue} (3)}(6x+3)^{\color{Blue} {(3-1)}}({\color{Blue} (1)}6x^{\color{Blue} {(1-1)}}+0) + ({\color{Blue} -\sin}(4x))({\color{Blue} (1)}4x^{\color{Blue} {(1-1)}})$

Therefore, the derivative is: $f'(x) = 18x(6x+3)^2 -4\sin(4x)$

Then, you plug $x=\frac{1}{2}$ into the derivative: $f'(\frac{1}{2}) = 18(\frac{1}{2})(6(\frac{1}{2})+3)^2 -4\sin(4(\frac{1}{2}))$

Therefore, the answer is: 320.4

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Example Question #403 : How To Find Rate Of Change

Find the slope at x=2 :

f(x)= (ln(2x))^2

Possible Answers:

Answer not listed

6.7

1.4

1.2

3.5

Correct answer:

1.4

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: f(x)= (ln(2x))^2

That is done by doing the following: $f'(x)= ln(2x) (\frac{2}{x})$

Then plug x=2 into the derivative: $f'(2)= ln(2({\color{Blue} 2})) (\frac{2}{{\color{Blue} 2}})$

Therefore, the answer is: 1.4

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Example Question #404 : How To Find Rate Of Change

Find the slope at x=1 :

f(x)= 5 ln (5x^2-3x+1)

Possible Answers:

13.5

Answer not listed.

10.2

7.4

9.3

Correct answer:

9.3

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: f(x)= 5 ln (5x^2-3x+1)

That is done by doing the following: $f'(x)=\frac{50x+ 15}{5x^2+3x-1}$

Then plug x=1 into the derivative: $f'(1)=\frac{50({\color{Blue}1 })+ 15}{5({\color{Blue} 1})^2+3({\color{Blue} 1})-1}$

Therefore, the answer is: 9.3

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Example Question #405 : How To Find Rate Of Change

Find the slope at x=1 :

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

Possible Answers:

Answer not listed.

6.5

23.5

10.2

5.4

Correct answer:

5.4

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: $f(x)= e^{3x^2+2x+1}$

That is done by doing the following: $f'(x)= e^{3x^2+2x+1} (6x + 2)$

Then plug x=1 into the derivative: $f'(0)= e^{3({\color{Blue} 0})^2+2({\color{Blue} 0})+1} (6({\color{Blue} 0}) + 2)$

Therefore, the answer is: 5.4

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Example Question #495 : Rate

Find the slope at x=0 :

$f(x)= \cos(\frac{1}{4}x^4 +3x)$

Possible Answers:

Answer not listed.

Correct answer:

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: $f(x)= \cos(\frac{1}{4}x^4 +3x)$

That is done by doing the following: $f'(x)= -\sin(\frac{1}{4}x^4 +3x) (x^{3} +3)$

Then plug x=0 into the derivative: $f'(0)= -\sin(\frac{1}{4}({\color{Blue} 0})^4 +3({\color{Blue} 0})) ({\color{Blue} 0}^{3} +3)$

Therefore, the answer is:

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Example Question #496 : Rate

Find the slope at x=1 :

$f(x)= 2x^2 + \tan x - (2x-4)^2$

Possible Answers:

9.7

Answer not listed.

2.4

16.7

15.4

Correct answer:

15.4

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative of that function.

In this case, we must find the derivative of the following: $f(x)= 2x^2 + \tan x - (2x-4)^2$

That is done by doing the following: $f'(x)= \sec^2x - 4x+16$

Then plug x=1 into the derivative: $f'(x)= \sec^2({\color{Blue} 1}) - 4({\color{Blue} 1})+16$

Therefore, the answer is: 15.4

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #482 : Rate

Example Question #483 : Rate

Example Question #3311 : Calculus

Example Question #401 : How To Find Rate Of Change

Example Question #402 : How To Find Rate Of Change

Example Question #403 : How To Find Rate Of Change

Example Question #404 : How To Find Rate Of Change

Example Question #405 : How To Find Rate Of Change

Example Question #495 : Rate

Example Question #496 : Rate

All Calculus 1 Resources

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