Applications of Differentiation

Question

Consider the function f(x)= 2x^3 - 6x^2 - 48x + 7 on the interval [-4,8]. Find the average or mean slope of the function on this interval.

Average slope: 24

By the Mean Value Theorem, we know there exists at least one value c in the open interval (-4,8) such that f'(c) is equal to this mean slope. List all values c that work.

Values of c: 

Answer 1

f'(x) = 6x² - 12x - 48; f(8) = 2·8³ - 6·8² - 48·8 + 7 = 263; f(- 4) = 2(- 64) - 6·16 + 48·4 + 7 = - 25

By Mean Value Theorem f'(c) = (f(8) - f(- 4))/(8 - (- 4));

6c² - 12c - 48 = (263 - (- 25)/12;

6c² - 12c - 48 = 24; c² - 2c - 12 = 0; c = 1 ± √13; c = 1 - √13 = - 2.61 or c = 1 + √13 = 4.61