rolle's theorem

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November 10th 2011, 05:47 PM
delgeezee

rolle's theorem

rolle's theorem
f must be continuous on a closed interval [a,b]
differential on (a,b) with f(a)=f(b)
Then, there is atleast one point C in (a,b) such that f'(c)=0

I am suppose to determine if rolle's theorem applies to the fuction and solve
$f(x)=1-x^{2/3}$ in the inverval [-1,1]

Mathab says the assumptions for the Theorem have not been satisfied. I am under the impression they are satisfied. I was wonderifing if someone could provide and explanation as to why the theorem has not been satisfied.
November 10th 2011, 07:10 PM
Prove It

Re: rolle's theorem

Quote:

Originally Posted by delgeezee

rolle's theorem
f must be continuous on a closed interval [a,b]
differential on (a,b) with f(a)=f(b)
Then, there is atleast one point C in (a,b) such that f'(c)=0

I am suppose to determine if rolle's theorem applies to the fuction and solve
$f(x)=1-x^{2/3}$ in the inverval [-1,1]

Mathab says the assumptions for the Theorem have not been satisfied. I am under the impression they are satisfied. I was wonderifing if someone could provide and explanation as to why the theorem has not been satisfied.

f(-1) = 0 and f(1) = 0, so Rolle's Theorem is satisfied.
November 10th 2011, 07:20 PM
Deveno

Re: rolle's theorem

is f(x) differentiable on (-1,1)? let's see:

$f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$

hmm....looks like we have a problem at x = 0.....
November 10th 2011, 07:21 PM
pickslides

Re: rolle's theorem

Is $f(-1)= f(1)$ ?
November 10th 2011, 07:27 PM
Prove It

Re: rolle's theorem

Quote:

Originally Posted by pickslides

Is $f(-1)= f(1)$ ?

It certainly is...

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

and

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

However, Deveno is right about the function not being differentiable in this entire region. So I was wrong about Rolle's Theorem being satisfied :)