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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #522 : Other Differential Functions

Consider a line tangent to the function f(x) at point (1,3) . If this line also passes through point (4,9) , then the following must be true:

Possible Answers:

f'(1)=-2

f'(4)=2

$f'(4)=\frac{1}{2}$

f'(1)=2

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

Correct answer:

f'(1)=2

Explanation:

The slope of the tangent line, , is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:

f'(x)=m

We can find the slope of the tangent line in this problem using rise over run:

$m=\frac{f(b)-f(a)}{b-a}$

$m=\frac{9-3}{4-1}=2$

Therefore:

f'(1)=2

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Example Question #711 : Differential Functions

Consider a line tangent to the function f(x) at point (4,5) . If this line also passes through point (2,13) , then the following must be true:

Possible Answers:

f'(2)=-4

$f'(5)=\frac{1}{4}$

f'(4)=-4

f'(4)=4

f'(5)=-4

Correct answer:

f'(4)=-4

Explanation:

The slope of the tangent line, , is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:

f'(x)=m

We can find the slope of the tangent line in this problem using rise over run:

$m=\frac{f(b)-f(a)}{b-a}$

$m=\frac{13-5}{2-4}=-4$

Therefore:

f'(4)=-4

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Example Question #524 : Other Differential Functions

Consider a line tangent to the function f(x) at point (2,9) . If this line also passes through point (4,5) , then the following must be true:

Possible Answers:

f'(4)=-2

f'(2)=2

f'(2)=-2

$f'(2)=-\frac{1}{2}$

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

Correct answer:

f'(2)=-2

Explanation:

The slope of the tangent line, , is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:

f'(x)=m

We can find the slope of the tangent line in this problem using rise over run:

$m=\frac{f(b)-f(a)}{b-a}$

$m=\frac{5-9}{4-2}=-2$

Therefore:

f'(2)=-2

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Example Question #525 : Other Differential Functions

Consider a line tangent to the function f(x) at point (3,1) . If this line also passes through point (8,8) , then the following must be true:

Possible Answers:

$f'(1)=\frac{5}{7}$

$f'(3)=\frac{5}{7}$

$f'(3)=\frac{7}{5}$

$f'(8)=\frac{7}{5}$

$f'(8)=\frac{5}{7}$

Correct answer:

$f'(3)=\frac{7}{5}$

Explanation:

The slope of the tangent line, , is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:

f'(x)=m

We can find the slope of the tangent line in this problem using rise over run:

$m=\frac{f(b)-f(a)}{b-a}$

$m=\frac{8-1}{8-3}=\frac{7}{5}$

Therefore:

$f'(3)=\frac{7}{5}$

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Example Question #526 : Other Differential Functions

Consider a line tangent to the function f(x) at point (19,25) . If this line also passes through point (2,11) , then the following must be true:

Possible Answers:

$f'(2)=\frac{17}{14}$

$f'(19)=\frac{17}{14}$

$f'(19)=\frac{14}{17}$

$f'(14)=\frac{19}{17}$

$f'(2)=\frac{14}{17}$

Correct answer:

$f'(19)=\frac{14}{17}$

Explanation:

The slope of the tangent line, , is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:

f'(x)=m

We can find the slope of the tangent line in this problem using rise over run:

$m=\frac{f(b)-f(a)}{b-a}$

$m=\frac{11-25}{2-19}=\frac{-14}{-17}=\frac{14}{17}$

Therefore:

$f'(19)=\frac{14}{17}$

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Example Question #527 : Other Differential Functions

Consider a line tangent to the function f(x) at point $(\pi,5e)$ . If this line also passes through point $(2e,4\pi)$ , then the following must be true:

Possible Answers:

$f'(\pi)=\frac{4\pi-5e}{2e-\pi}$

$f'(2e)=\frac{2e-\pi}{4\pi-5e}$

$f'(2e)=-\frac{\pi}{e}$

$f'(\pi)=-\frac{e}{\pi}$

$f'(\pi)=-\frac{\pi}{e}$

Correct answer:

$f'(\pi)=\frac{4\pi-5e}{2e-\pi}$

Explanation:

The slope of the tangent line, , is equal to the slope of the function where it is defined. Furtheremore, the derivative of the function at this point is equal to the value of this slope:

f'(x)=m

We can find the slope of the tangent line in this problem using rise over run:

$m=\frac{f(b)-f(a)}{b-a}$

$m=\frac{4\pi-5e}{2e-\pi}$

Therefore:

$f'(\pi)=\frac{4\pi-5e}{2e-\pi}$

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Example Question #711 : Differential Functions

Given that $\frac{dy}{dx}=x^2$ and y(1)=2 , use Euler's method to approximate y(4) using three steps.

Possible Answers:

Correct answer:

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{4-1}{3}=1$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

y_0=2

y_1=y_0+1(1^2)=2+1=3

y_2=3+1(2^2)=7

y_3=7+1(3^2)=16

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Example Question #712 : Differential Functions

Given that $\frac{dy}{dx}=tan(x^3+5)$ and y(4)=7 , use Euler's method to approximate y(6) using three steps.

Possible Answers:

6.54

6.11

6.80

6.92

Correct answer:

6.54

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{6-4}{3}=\frac{2}{3}$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

y_0=7

$y_1=7+\frac{2}{3}tan(4^3+5)=6.92$

$y_2=6.92+\frac{2}{3}tan((\frac{14}{3})^3+5)=6.80$

$y_3=6.80+\frac{2}{3}tan((\frac{16}{3})^3+5)=6.54$

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Example Question #713 : Differential Functions

Given that $\frac{dy}{dx}=e^{\frac{5}{x}}$ and y(3)=12 , use Euler's method to approximate y(4) using three steps.

Possible Answers:

13.76

16.55

14.88 17.97

15.25

Correct answer:

16.55

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{4-3}{3}=\frac{1}{3}$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

y_0=12

$y_1=12+\frac{1}{3}e^{\frac{5}{3}}=13.76$

$y_2=13.76+\frac{1}{3}e^{\frac{5}{\frac{10}{3}}}=15.25$

$y_3=15.25+\frac{1}{3}e^{\frac{5}{\frac{11}{3}}}=16.55$

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Example Question #1750 : Calculus

Given that $\frac{dy}{dx}=e^{x^3-1}$ and y(0)=1.8 , use Euler's method to approximate y(0.3) using three steps.

Possible Answers:

2.032

1.8

2.259

1.874

1.911

Correct answer:

1.911

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{0.3-0}{3}=0.1$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

y_0=1.8

$y_1=1.8+0.1(e^{0^3-1})=1.837$

$y_2=1.837+0.1(e^{0.1^3-1})=1.874$

$y_3=1.874+0.1(e^{0.2^3-1})=1.911$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #522 : Other Differential Functions

Example Question #711 : Differential Functions

Example Question #524 : Other Differential Functions

Example Question #525 : Other Differential Functions

Example Question #526 : Other Differential Functions

Example Question #527 : Other Differential Functions

Example Question #711 : Differential Functions

Example Question #712 : Differential Functions

Example Question #713 : Differential Functions

Example Question #1750 : Calculus

All Calculus 1 Resources

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