Mean Value Theorem

Question

Let f(x) = (x^2 - 6x)e^x on [0,6]

Find c such as f'(c) = 0, or if Rolle's Theorem does not apply, enter DNE

c =

Answer 1

For f(x) = (x² - 6x)e^x on [0,6], then f(0) = 0 and f(6) = 0 therefore Rolle's theorem applies which means there is some value between x = 0 and x = 6 for which f '(x) = 0

To find that value of "x", (or the "c" in Rolle's Theorem) take the derivative. To do so, use the product rule (u•v)' = u'v + uv'

u = x² - 6x

u' = 2x - 6

v = e^x

v' = e^x

So f '(x) = (2x - 6)e^x + (x² - 6x)e^x

(2x - 6)e^x + (x² - 6x)e^x = 0

e^x[(2x - 6) + (x² - 6x)] = 0

e^x(x² - 4x - 6) = 0

Setting each factor equal to zero:

e^x = 0 (no solution)

or x² - 4x - 6 = 0 and using the quadratic formula:

x = [4 ± √(16 - 4(1)(-6)]/2 = (4 ± √40)/2 = (4 ± 2√10)/2 = 2 ± √10

The value that is between 0 and 6 is 2 + √10 (approx 5.16) therefore c = 2 + √10