What are the critical values, if any, of # f(x)=x^3 + 6x^2 − 135x#?

1 Answer
Nov 19, 2015

Answer:

#-9# and #5#.

Explanation:

A critical number for a function #f#, is a number in the domain of #f# at which either the derivative of #f# is #0# or it fails to exist.

For # f(x)=x^3 + 6x^2 − 135x#, the domain is #(-oo,oo)# and

#f'(x) = 3x^2+12x-136#.

This #f'(x)# exists for all numbers.

#f'(x)=0# at #-9# and at #5#, both of which are in the domain of #f#, so both are critical numbers for #f#.

To solve #f'(x)=0#, use the quadratic formula or factor:

#3x^2+12x-136 = 3(x^2+4x-45) = 3(x+9)(x-5)#