The function f(x)=(3x+9)e^-3x has one critical number. Find it.

Answer 1

Take the derivative and set it equal to zero to find critical points.

To the derivative, use the product rule:

f'(x) = u'v + v'u where u = 3x + 9, u' = 3, v = e^-3x, and v' = -3e^-3x

f'(x) = 3e^-3x + (-3e^-3x)(3x + 9)

factor out 3e^-3x to get f'(x) = 3e^-3x(1 - (3x + 9))

f'(x) = 3e^-3x(1 - 3x - 9))

f'(x) = 3e^-3x(-8 - 3x)

Set f' = 0 so: 3e^-3x(-8 - 3x) = 0

That means 3e^-3x = 0 and (-8 - 3x) = 0

There is no solution for 3e^-3x = 0

For (-8 - 3x) = 0, -8 = 3x or x = -8/3

The critical number is x = -8/3.