How do you find the critical points f(x)= 2x^3 + 3x^2 - 36x + 5?

1 Answer
Aug 10, 2015

(-3,86) and (2, -39)

Explanation:

From Wikipedia: a critical point or stationary point of a differentiable function of a single real variable, f(x), is a value x_0 in the domain of f where its derivative is 0: f′(x_0) = 0

Thus, to find the critical points of f(x) = 2x^3+3x^2-36x+5, we first need to compute f'(x) then find all the x-values such that f'(x)=0.

f(x) = 2x^3+3x^2-36x+5
f'(x) = 6x^2+6x-36x

When f'(x)=0:
6x^2+6x-36x=0
x^2+x-6x=0
(x+3)(x-2)=0
x=-3 or x=2

x=-3, y=2(-3)^3+3(-3)^2-36(-3)+5=86
x=2, y=2(2)^3+3(2)^2-36(2)+5=-39

Thus, critical points are (-3,86) and (2, -39) You could conjecture that the first is a maximum point and the second a minimum, but you'll need the second derivative test to prove that in totality.