Question

Absolute minimum and maximum values of $f(x)=2x^{3}-3x^{2} -12x+1$ on the interval [-2,3]

Answer 1

Step-by-step explanation:

$f'(x)=6x^{2}-6x-12$

So f'(x)=0:

x^2 - x - 2 = 0

(x-2)(x+1)=0

x=-1, 2

So we need to find the value of f at those critical points and also at the endpoints of the interval

f(-1)=-2-3+12+1=8

f(2)=16-12-24+1=-19

f(-2)=-16-12+24+1=-3

f(3)=54-27-36+1=-8

so the max is 8 and the min is -19