Given f x 2x3 3x2 36 x 14 Find the critical numbers Find

Given: f (x) = 2x^3 - 3x^2 - 36 x +14 Find the critical numbers. Find the relative (local) extrema.

Solution

f(x)=2x^3-3x^2-36x+14

f\'(x)=6x^2-6x-36

for critical points we put f\'(x)=0

6(x^2-x-6)=0

x(x-3)-2(x-3)=0

(x-3)(x+2)=0

x=3,-2 are critical points

now f(3)=2(3^3)-3(3^2)-36*3+14=-67

f(-2)=2(-2)^3-3(-2)^2-36*-2+14=-16-12+72+14=58

local extrema is at (-2,58)