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Calculus 1 : How to find differential functions

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Example Questions

Example Question #708 : Differential Functions

As per the mean value theorem, there exists at least one value x=c within the interval (1,b) such that f'(c)=-4 . For the function, interval, and derivative value f(x)=x^2-e^x , (1,b) , , find a value of that validates the mean value theorem.

Possible Answers:

3.33

2.84

4.75

5.11

-4.51

Correct answer:

2.84

Explanation:

The mean value theorem states that for a planar arc passing through a starting and endpoint (a,b);a<b , there exists at a minimum one point, , within the interval (a,b) for which a line tangent to the curve at this point is parallel to the secant passing through the starting and end points.

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Using our function, interval, and derivative definitions, f(x)=x^2-e^x , (1,b) , f'(c)=-4 , we'll in turn wish to solve for to validate the mean value theorem:

$-4=\frac{(b^2-e^b)-(1^2-e^1)}{b-1}$

Solving for using a calculator gives the solution:

b=-4.51,2.84

To enable the creation of an interval, elect the value greater than a:

b=2.84

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Example Question #709 : Differential Functions

As per the mean value theorem, there exists at least one value x=c within the interval (-2,b) such that f'(c)=25 . For the function, interval, and derivative value f(x)=x^3+e^x , (-2,b) , , find a value of that validates the mean value theorem.

Possible Answers:

-1.08

0.91

3.31

4.27

3.99

Correct answer:

4.27

Explanation:

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Using our function, interval, and derivative definitions, f(x)=x^3+e^x , (-2,b) , f'(c)=25 , we'll in turn wish to solve for to validate the mean value theorem:

$25=\frac{(b^3+e^b)-((-2)^3+e^{-2})}{b-(-2)}$

Solving for using a calculator gives the solution:

b=-3.68,4.27

Elect the value that is greater than a to complete the interval:

b=4.27

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Example Question #710 : Differential Functions

As per the mean value theorem, there exists at least one value x=c within the interval (-3,b) such that f'(c)=0 . For the function, interval, and derivative value $f(x)=x^3+e^{x^2}$ , (-3,b) , f'(c)=0 , find a value of that validates the mean value theorem.

Possible Answers:

2.66

3.00

3.38

4.12

1.91

Correct answer:

3.00

Explanation:

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Using our function, interval, and derivative definitions, $f(x)=x^3+e^{x^2}$ , (-3,b) , f'(c)=0 , we'll in turn wish to solve for to validate the mean value theorem:

$0=\frac{(b^3+e^{b^2})-((-3)^3+e^{(-3)^2})}{b-(-3)}$

Solving for using a calculator gives the solution:

b=3.00

Which is, as expected, greater than to allow the creation of an interval.

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Example Question #521 : How To Find Differential Functions

As per the mean value theorem, there exists at least one value x=c within the interval (1.5,b) such that f'(c)=10 . For the function, interval, and derivative value f(x)=x^3-3x^2-4x , (1.5,b) , , find a value of that validates the mean value theorem.

Possible Answers:

4.85

3.35

1.99

-3.35

5.15

Correct answer:

4.85

Explanation:

Meanvaluetheorem

In other words, if one were to draw a straight line through these start and end points, one could find a point on the curve where the tangent would have the same slope as this line.

$f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{Rise}{Run};a<c<b$

Using our function, interval, and derivative definitions, f(x)=x^3-3x^2-4x , (1.5,b) , f'(c)=10 , we'll in turn wish to solve for to validate the mean value theorem:

$10=\frac{(b^3-3b^2-4b)-(1.5^3-3(1.5)^2-4(1.5))}{b-1.5}$

Solving for using a calculator gives the solution:

b=-3.35,4.85

Although there are two solutions, keep in mind that to satisfy the mean value theorem, we find a value to define the interval (1.5,b) , and as such b should be greater than 1.5

b=4.85

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Example Question #522 : How To Find 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:

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

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

f'(1)=-2

f'(4)=2

f'(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 #523 : How To Find 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'(4)=-4

f'(2)=-4

f'(5)=-4

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

f'(4)=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 : How To Find 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'(2)=-2

f'(4)=-2

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

f'(2)=2

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

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 : How To Find 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:

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

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

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

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

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

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 : How To Find 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:

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

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

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

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

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

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 : How To Find 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:

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

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

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

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

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

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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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #708 : Differential Functions

Example Question #709 : Differential Functions

Example Question #710 : Differential Functions

Example Question #521 : How To Find Differential Functions

Example Question #522 : How To Find Differential Functions

Example Question #523 : How To Find Differential Functions

Example Question #524 : How To Find Differential Functions

Example Question #525 : How To Find Differential Functions

Example Question #526 : How To Find Differential Functions

Example Question #527 : How To Find Differential Functions

All Calculus 1 Resources

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