Question

i have this function: f(x)=x^1/3+1, i got the derivative as being 1/3x^-2/3, now in this form it is still defined at zero,but in this form it is not: (1/3)/(x^2/3). The only critical number for this is 0. Will there be and relative extrema.

Answer 1

The critical number is 0 because the derivaitve is undeifnined there right?

Answer 2

critical numbers are at f' = 0 and at f' = undefined

Answer 3

so, does this function have relative extrema

Answer 4

look for sign changes in f'; if they go from left to right:

+ to -, you got a change in slope like going over a hill

- to +, you got a change in slope like going into a bowl.

if its the same sign on each side, its most like ly an inflection

Answer 5

the cbrt function +1 has no min or max; just an inflection

Answer 6

I know the critical number is 0, cause this derivative is undefined there, what will be the intervals that i have to test in regards to this function?

Answer 7

since x^2 in the derivative always makes this a positive number we get:

<..........0............>
+ +

the slope is always increaseing from left to tight

Answer 8

left to right lol

Answer 9

when it hits 0, its a vertical slope

Answer 10

but i dont have x^2

Answer 11

x^(1/3) +1

1
--------- ; right?
3 cbrt(x^2)

Answer 12

yes

Answer 13

then id say you got an x^2 :)

Answer 14

no matter what you use for x, itll pop out a positive value

Answer 15

right, thanks

Answer 16

youre welcome :)