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?