Question

Find the absolute maximum and absolute minimum values of f on the given interval.
f(x) = (x^2 - 1)^3
[-1, 6] will fan and medal!!!

Answer 1

First step: Differentiate f
Second step: Find critical numbers.

Answer 2

Third step: Is a matter of pluggin' into the original function

Answer 3

Evaluate f(x) at the end points, x = -1 and x = 6 along with f(x) values at the critical point to determine the absolute max and min.

Answer 4

f'(x)=(2x)^3
f'(x)=(6x)^2
@freckles

Answer 5

To differentiate f(x)=(x^2-1)^3
you need to chain rule, power rule, difference rule, power rule again, constant rule

Answer 6

\[f'(x)=3(x^2-1)^{3-1} \cdot (x^2-1)'\]

Answer 7

to find the critical number, do I make f'(x) = 0 @freckles

Answer 8

you set f'=0 and solve for x
have you finished finding f' yet?

Answer 9

f'(x)= 6x^5-12x^3+6x

Answer 10

well I would have left it in factored form but okay

Answer 11

\[f'(x)=3(x^2-1)^{3-1}(x^2-1)' =3(x^2-1)^2(2x) \\ \text{ set} f'=0 \\ 3(x^2-1)^2(2x)=0 \text{ when } x=?\]

Answer 12

x=0, x=1

Answer 13

@freckles

Answer 14

there is x=-1 but that is endpoint anyways

Answer 15

now plug into f

Answer 16

plug the endpoints and the critical numbers which occured between the endpoints into f

Answer 17

\[f(-1)=? \\f(0)=? \\ f(1)=? \\f(6)=?\]

Answer 18

f(-1)=-8
f(0)=-1
f(6)=42875
@freckles

Answer 19

how id you get -8 for the first one
and then you find f(1) too?

Answer 20

f(-1)=(-1^2)^3=-8
f(1)=0

Answer 21

\[f(-1)=((-1)^2-1)^3=(1-1)^3=0\]

Answer 22

both f(-1) and f(1) is zero

Answer 23

so the outputs we got were 0,-1,42875
-
-1 is the lowest value (so that is the absolute min on the interval)
42875 is the highest value (so this is the absolute max on the interval)