Answer to Question #74194 in Calculus for yewande folayan

Question

Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a single category list the points in increasing order in x value and enter N in any blank that you don't need to use.

f(x)=4 (x^2 - 1)^3, \ [-1,2]

Absolute maxima, absolut minima

Answer 1

f(x)=4(x^2-1)^3,[-1,2]
Find the first derivative with respect to x
f^' (x)=(4(x^2-1)^3 )^'=4(3) (x^2-1)^2 (x^2-1)^'=12(2x) (x^2-1)^2=
=24x(x^2-1)^2
Find the critical point(s)
f^' (x)=0=>24x(x^2-1)^2=0
24x(x-1)^2 (x+1)^2=0
x_1=-1,x_2=0,x_3=1
To find the absolute maximum and absolute minimum, then, we evaluate f at the critical points and on the endpoints of the interval:
f(-1)=4((-1)^2-1)^3=0
f(0)=4((0)^2-1)^3=-4
f(1)=4((1)^2-1)^3=0
f(2)=4((2)^2-1)^3=108
Therefore, f achieves its absolute minimum of -4 at x=0 and its absolute maximum of 108 at x=2.

Absolute maxima 108

Absolute minima -4