Consider the function f(x)=2x^3-9x^2-108x+7 on the interval [-6,10]

Question

The average or mean slope of the function on this interval is equal to ____

By the mean value theorem, we known there exists a c in the open interval (-6,10) such that f'(c) is equal to this mean slope. For this problem, there are two values of c that work:

The smaller one c1=

The larger one c2=

Answer 1

Given function is f(x) = 2x³ - 9x² - 108x + 7.

f(-6) = 2(-6)³ - 9(-6)² - 108(-6) + 7 = -101

f(10) = 2(10)³ - 9(10)² - 108(10) + 7 = 27

Mean slope = (f(10) - f(-6))/(10 - (-6)) = (27 - (-101))/(16) = 8

Differentiating f(x) with respect to x, we get

f'(x) = d/dx (2x³ - 9x² - 108x + 7) = 6x² - 18x - 108

Setting f'(c) = mean slope, we get

f'(c) = 8

6c² - 18c - 108 = 8

3c² - 9c - 58 = 0

3(c² - 3c) - 58 = 0

3(c² - 3c + 9/4 - 9/4) - 58 = 0

3(c² - 3c + 9/4) - 27/4 - 58 = 0

3(c - 3/2)² - 259/4 = 0

(c - 3/2)² = 259/12

c = 3/2 ± √(259/12)

c = (3 ± √(259/3))/2

So, the smaller c is c₁ = (3 - √(259/3))/2 ≈ -3.146

and the larger c is c₂ = (3 + √(259/3))/2 ≈ 6.146

and -6 < c₁ < c₂ < 10.