What are the critical points of f(x)=4x^3+8x^2+6x-2?

1 Answer
Feb 25, 2016

There is no critical point for the given function

Explanation:

f'(x) = 12x^2+8x+6

for a critical point f'(x)=0
however the discriminant of 12x^2+8x+6
color(white)("XXX")b^2-4ac=8^2-4(12)(6) < 0
so there are no Real values of x for which the function has a zero slope (i.e. there are no critical points).

A graph of the function looks like:
graph{4x^3+8x^2+6x-2 [-2.87, 2.605, -4.936, -2.202]}