Rolle's Theorem Problem - # 2

[Jason76]

\(\displaystyle f(x) = \sqrt{x} - \dfrac{1}{9}x\)

\(\displaystyle f(x) = x^{1/2} - \dfrac{1}{9}x\)

It is continuous and differentiable across all real numbers.

\(\displaystyle [0,81]\)
\(\displaystyle f(0) = \sqrt{0} - \dfrac{1}{9}(0) = 0\)

\(\displaystyle f(81) = \sqrt{81} - \dfrac{1}{9}(81) = 0\)


\(\displaystyle f'(x) = \dfrac{1}{2}x^{-1/2} - \dfrac{1}{9}\)

\(\displaystyle \dfrac{1}{2}x^{-1/2} - \dfrac{1}{9} = 0\)

\(\displaystyle \dfrac{1}{2}x^{-1/2} = \dfrac{1}{9}\)

\(\displaystyle x^{-1/2} = \dfrac{1}{9}(\dfrac{2}{1})\)

\(\displaystyle x^{-1/2} = \dfrac{2}{9}\)

\(\displaystyle [x^{-1/2}]^{-2} = [\dfrac{2}{9}]^{-2}\)

\(\displaystyle x = (\dfrac{2}{9})^{-2}\) What would this lead to?

 

[stapel]

\(\displaystyle x = (\dfrac{2}{9})^{-2}\) What would this lead to?

Dunno. What question are you supposed to be answering?

 

[Jason76]

Dunno. What question are you supposed to be answering?

\(\displaystyle x = (\dfrac{2}{9})^{-2}\) What is \(\displaystyle (\dfrac{2}{9})^{-2}\) ?

 

[DrPhil]

\(\displaystyle x = (\dfrac{2}{9})^{-2}\) What is \(\displaystyle (\dfrac{2}{9})^{-2}\) ?

20.25

What is Rolle's Theorem, and what are you trying to find?

 

[Jason76]

Answer comes out to \(\displaystyle \dfrac{81}{4}\) which makes sense cause

\(\displaystyle [\dfrac{2}{9}]^{2} = \dfrac{4}{81}\) so, naturally the negative of the exponent would come out to the reciprocal. Anyone agree?

 

[DrPhil]

Answer comes out to \(\displaystyle \dfrac{81}{4}\) which makes sense cause

\(\displaystyle [\dfrac{2}{9}]^{2} = \dfrac{4}{81}\) so, naturally the negative of the exponent would come out to the reciprocal. Anyone agree?

Yes, as I said, 20.25.

But what is Rolle's Theorem, and what are you trying to show? Have you shown it?