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

1 Answer
Aug 10, 2015

Answer:

#(-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.