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The table above gives selected values for the differentiable function f. in which of the following intervals must there be a number c such that f’(c)=2 PHOTO ATTACHED

PLEASE HELP The table above gives selected values for the differentiable function f in which of the following intervals must there be a number c such that fc2 P class=

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Answer:

C.

Step-by-step explanation:

We need to apply Mean Value theorem here.

We just need to calculate the average rate of change for each interval.

A) x=0 to x=4

We have the points (0,8) and (4,0) are the points that correspond to the endpoints of the given interval.

(8-0)/(0-4)

8/-4

-2

Not this choice

B) x=4 to x=8

We have the points (4,0) and (8,2) are the points that correspond to the endpoints of the given interval.

(2-0)/(8-4)

2/4

1/2

Not this choice

C) x=8 to x=12

We have the points (8,2) and (12,10) are the points that correspond to the endpoints of the given interval.

(10-2)/(12-8)

8/4

2

This one works.

D) x=12 to x=16

We have the points (12,10) and (16,1) are the points that correspond to the endpoints of the given interval.

(1-10)/(16-12)

-9/4

MrRoyal

The table is an illustration of mean value theorem

[tex]\mathbf{f'(c) =2}[/tex] on the interval (8,12)

If a function is differentiable, then:

[tex]\mathbf{f'(c) =\frac{f(b) - f(a)}{b - a}}[/tex]

(a) (0,4)

This gives

[tex]\mathbf{f'(c) =\frac{f(4) - f(0)}{4 - 0}}[/tex]

[tex]\mathbf{f'(c) =\frac{f(4) - f(0)}{4}}[/tex]

From the table, we have:

[tex]\mathbf{f'(c) =\frac{0 - 8}{4}}[/tex]

[tex]\mathbf{f'(c) =\frac{- 8}{4}}[/tex]

[tex]\mathbf{f'(c) =-2}[/tex]

(b) (4,8)

This gives

[tex]\mathbf{f'(c) =\frac{f(8) - f(4)}{8 - 4}}[/tex]

[tex]\mathbf{f'(c) =\frac{f(8) - f(4)}{4}}[/tex]

From the table, we have:

[tex]\mathbf{f'(c) =\frac{2-0}{4}}[/tex]

[tex]\mathbf{f'(c) =\frac{2}{4}}[/tex]

[tex]\mathbf{f'(c) =\frac{1}{2}}[/tex]

(c) (8,12)

This gives

[tex]\mathbf{f'(c) =\frac{f(12) - f(8)}{12 - 8}}[/tex]

[tex]\mathbf{f'(c) =\frac{f(12) - f(8)}{4}}[/tex]

From the table, we have:

[tex]\mathbf{f'(c) =\frac{10-2}{4}}[/tex]

[tex]\mathbf{f'(c) =\frac{8}{4}}[/tex]

[tex]\mathbf{f'(c) =2}[/tex]

Hence, [tex]\mathbf{f'(c) =2}[/tex] on the interval (8,12)

Read more about mean value theorems at:

https://brainly.com/question/3957181

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