Answer:
The critical points of [tex]f(x) = 2\cdot x^{3}-3\cdot x^{2}-36\cdot x[/tex] are [tex]x_{1} = 3[/tex] and [tex]x_{2} = -2[/tex]. {-2, 3}
Step-by-step explanation:
We must use First Derivative Test to determine all critical numbers of the function [tex]f(x) = 2\cdot x^{3}-3\cdot x^{2}-36\cdot x[/tex]
As first step we need its first and second derivatives:
First derivative
[tex]f'(x) = 6\cdot x^{2}-6\cdot x-36[/tex] (Eq. 1)
Now, we equalize (Eq. 1) to find all critical numbers and solve it afterwards:
[tex]6\cdot x^{2}-6\cdot x-36 = 0[/tex]
By Quadratic Formula all roots are found:
[tex]x_{1} = 3[/tex] and [tex]x_{2} = -2[/tex]
The critical points of [tex]f(x) = 2\cdot x^{3}-3\cdot x^{2}-36\cdot x[/tex] are [tex]x_{1} = 3[/tex] and [tex]x_{2} = -2[/tex]. {-2, 3}
