Consider the function f(x)=2x3−6x2−48x+9 on the interval [−4,7]. Find the average or mean slope of the function on this interval. Average slope = 8 equation editorEquation Editor By the Mean Value Theorem, we know there exists at least one c in the open interval (−4,7) such that f′(c) is equal to this mean slope. Find all values of c that work and list them (separated by commas) in the box below.

A) Mean Slope = 8

B) Values of C; (-2.215, 4.215)

Step-by-step explanation:

A) Formula for the mean slope over an interval [a, b] is given as;

Mean slope = [f(b) - f(a)]/(b - a)

Now, for this question, we have;

f(x) = 2x³ − 6x² − 48x + 4

Thus, the mean slope of this function on interval [−4,7] will be;

[f(7) - f(-4)]/(7 - (-4))

f(7) = 2(7³) − 6(7)² − 48(7) + 4

f(7) = 60

f(-4) = 2(-4³) − 6(-4)² − 48(-4) + 4

f(-4) = -28

Thus, the mean value is now ;

[f(7) - f(-4)]/(7 - (-4)) = [60 - (-28)]/11

= 88/11 = 8

B) We seek to verify the Mean Value Theorem for the function

f(x) = 2x³ − 6x² − 48x + 4

on the interval [−4,7]

The Mean Value Theorem, tells us that if f(x) is differentiable on a interval [a,b], then ∃ c ∈ [a,b] st:

f'(c) = (f(b) − f(a)) /b − a

So, Differentiating with respect to x, we have;

f'(x) = 6x² - 12x - 48

And we seek a value c ∈[−4,7] st:

f'(c) = [f(7) − f(−4)]/(7 −(−4))

f'(c) = 6c² - 12c - 48

from earlier, f(7) = 60 and f(-4) = -28

Thus;

6c² - 12c - 48 = (60 - (-28))/11

6c² - 12c - 48 = 88/11

6c² - 12c - 48 = 8

6c² - 12c - 48 - 8 = 0

6c² - 12c - 56 = 0

Using quadratic formula to solve this, we have the roots as;

c = -2.215 or 4.215

the answer to this question is 3/19

step-by-step explanation:

c

step-by-step explanation:

i think because what else can be a sample of every member