Answer to Question #107623 in Calculus for victoria

Question

Find the absolute maximum and minimum values of the following function over 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.

Answer 1

The closed interval method

To find the absolute maximum and minimum values of a continuous function $f$ on a closed interval $[a,b]$:

1. Find the values of $f$ at the critical numbers of $f$ in $(a,b).$

2. Find the values of $f$ at the endpoints of the interval.

3. The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.

$f(x)=6(x^2-1)^3, [-1, 2]$

1. Find the first derivative with respect to $x$

$$f'(x)=(6(x^2-1)^3)'=6(3)(x^2-1)^2(2x)=36x(x^2-1)^2$$

Find the critical number(s)

$$f'(x)=0=>36x(x^2-1)^2=0$$

$x_1=-1,x_2=0,x_3=1$

$f(-1)=6((-1)^2-1)^3=0$

$f(0)=6((0)^2-1)^3=-6$

$f(1)=6((1)^2-1)^3=0$

2.

$f(-1)=6((-1)^2-1)^3=0$

$f(2)=6((2)^2-1)^3=162$

3. Absolute maxima

$$x=2\ \ \ \ y=162$$

Absolute minima

$$x=0\ \ \ \ y=-6$$