Question

Critical values
Can someone help me find the critical values of the function f(x)=x^(2/3)-x please?

Answer 1

https://www.google.com/webhp?tab=ww&ei=4Y0bVLv5KsSsjAKkg4DYBg&ved=0CAMQ1S4#q=f(x)%3Dx%5E(2%2F3)-x+

Answer 2

hoped this helped

Answer 3

Well I know I can find them that way.. but I need to calculate them with derivatives and I am having trouble on this one.

Answer 4

did you use the power rule

Answer 5

Yes I did I got f(x)=(2/3)x^(-1/3)-1

Answer 6

derivative should be
\[\frac{2}{3} x^{-1/3} - 1 = 0\]

Answer 7

My problem is solving for that..

Answer 8

ok remember that "negative exponent" represents reciprocal
\[x^{-1/3} = \frac{3}{2}\]
flip both sides
\[x^{1/3} = \frac{2}{3}\]
raise to 3rd power to cancel cube root
\[x = (\frac{2}{3})^3\]

Answer 9

Oh thanks.. that makes so much more sense. I will try for the inflection points and can I ask you if I have trouble with that too?

Answer 10

sure , do same thing with 2nd derivative for inflection points

Answer 11

and did you just ignore the -1 then?

Answer 12

what is derivative of 1 ?

Answer 13

Nevermind I see what you did.. just one more question, isn't 0 a CV too?

Answer 14

nope , plug x=0 into derivative and you have -1 = 0 which is false

0 is just the inflection point

Answer 15

Well when you look at the graph of the function it is Decrease (-infinity,0),(8/27,Infinity) and looks to increase from (0,8/27)

Answer 16

correct

Answer 17

so that means it must have 2 CVS.. this is why I am so confused

Answer 18

Wait a second... could this function have a domain issue.. I dont think negative numbers are in the domain

Answer 19

sorry for late reply

ok the problem is that the function is not differentiable at x=0
http://www.wolframalpha.com/input/?i=plot+x%5E%282%2F3%29&a=%5E_Real

Answer 20

So that means it is not a critical value because it is not zero ?

Answer 21

it depends, you can have negative x values if you take 3rd root first then square

Answer 22

its not a critical value simply because it does not make derivative zero

Answer 23

So if I had to state where the function was increasing and decreasing would I just start with (0 or would I still do (-infinity, because that just confuses me because it technically does decrease

Answer 24

yes it does decrease for (-inf,0)

i guess i may be wrong, depends how you define CV
if its any point where there is change in increase/decrease then x=0 is a CV but it is not a CV in the sense that its a local min/max

sorry for the confusion :{

Answer 25

It's fine! Our teacher told us where there is a sharp point it is not a critical value I think.. But I will treat it as one and just bring it up to him in class.

Answer 26

\(c\) is a critical value for \(f\) if \(f'(c)=0\) or \(f'(c)\) does not exist and \(c\) is in the domain of the original function

Answer 27

thank you @Zarkon

Answer 28

Oh, thanks Zarkon. That makes a lot of sense now