Expert: Paul Klarreich - 10/7/2005

Question
Hi!
I hope I'm finding you in the best of health.
Find the values of c that satisfy Rolle's Theorem for f(x)=x^(2/3) - x^(1/3) on the interval [0,1].
I got (2/3)c^(-1/3) - (1/3)c^(-2/3)=0.
According to my answer key, c=1/8. But how do you solve the above equation for c and get that?
Thank you so much for your help.

Answer
Hi, Jeff,

You wrote:
Hi!
I hope I'm finding you in the best of health.
==== I thank you for your good wishes. =======

Find the values of c that satisfy Rolle's Theorem for

f(x)=x^(2/3) - x^(1/3) on the interval [0,1].

I got (2/3)c^(-1/3) - (1/3)c^(-2/3)=0.
============================
Yes, that is correct.  That is [VIEW IN FIXED FONT]
  2         1
------- - -------- = 0
3c(1/3)   3c(2/3)

Now write  W = c^(1/3), with c = W^3.  That should simplify things considerably.
2     1
--- - ---- = 0
3W    3W^2

2     1
--- = ----
3W    3W^2

6W^2 = 3W

Which gives  W = 1/2 and W = 0, which gives c = 1/8 and c = 0, but as I recall, Rolle's theorem requires that the solutions be in the OPEN interval, so c = 0 is no good.