Mean Value Theorem

Question

Consider the function  on the interval . Find the average or mean slope of the function on this interval.  By the Mean Value Theorem, we know there exists a  in the open interval  such that  is equal to this mean slope. For this problem, there are two values of  that work.The smaller one is   and the larger one is   

Answer 1

First, find the average slope of f(x) on the interval [-4, 7].

This is the slope between endpoints (-4, f(-4)) and (7, f(7)).

f(-4) = 2(-4)³ -12(-4)² - 30(-4) + 1 = -199

f(7) = 2(7)³ -12(7)² - 30(7) + 1 = -111

The average slope is m = (-111 - (-199)) / (7 - (-4)) = 88/11 = 8

Find c such that the derivative f '(c) = 8.

f '(x)= 6x² -24x - 30, so we need to solve the quadratic equation 6x² -24x - 30 = 8.

It is convenient to use the completing of square method here.

6x² -24x = 38; 6(x² - 4x) = 38; 6(x² - 4x + 4) = 38 + 24;

6(x - 2)² = 62; (x - 2)² = 62/6 = 31/3

x - 2 = ±√(31/3)

x = 2 ± √(31/3)

x₁ = 2 - √(31/3) ≈ -1.21455 x₂ = 2 + √(31/3) ≈ 5.21455

Both values are in the interval (-4, 7).