Question

Photo attached

Answer 1

what did you get for the derivative?

Answer 2

I got 2x^(-1/3)-2

Answer 3

yeah but that negative exponent doesn't help your cause

Answer 4

\[\frac{2}{\sqrt[3]{x}}-2\]

Answer 5

one critical point is clearly 0, because this is undefined when \(x=0\)

Answer 6

I see. so theres no answer?

Answer 7

yes, you still have to set it equal to zero and solve

Answer 8

you have to find the critical point, which are where the derivative is either equal to zero or undefined

Answer 9

so one critical point is \(x=0\)

Answer 10

for the other, set
\[\frac{2}{\sqrt[3]{x}}-2=0\] and solve for \(x\)

Answer 11

x=\[\sqrt[3]{1}\]?

Answer 12

that what I got. @satellite73

Answer 13

yeah , 1 is a critical point as well

Answer 14

now you have to check
\[f(-1),f(0),f(1)\] biggest is the max, smallest is the min

Answer 15

How do i check ?

Answer 16

f(-1) , put -1 in to the equation that given?

Answer 17

3(-1)^(2/3)-2(-1)

Answer 18

@satellite73
im still stuck with this question could you give me help