Rolle's Theorem - # 2

[Jason76]

Verify Rolle's Theorem and then check for values of \(\displaystyle x\) that satisfy it.

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

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

1st check: The function is differentiable and continuous over the interval, cause it's a polynomial.

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

\(\displaystyle f(81) = (81)^{1/2} - \dfrac{1}{9}(81) = 0\)

2nd check

\(\displaystyle f(0) = f(81)\)

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

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

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

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

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

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

\(\displaystyle x = \) :?:

 

[HallsofIvy]

Verify Rolle's Theorem and then check for values of \(\displaystyle x\) that satisfy it.

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

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

1st check: The function is differentiable and continuous over the interval, cause it's a polynomial.

Yes, the function is diffeferentiable and continuous but it is not a polynomial.

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

\(\displaystyle f(81) = (81)^{1/2} - \dfrac{1}{9}(81) = 0\)

2nd check

\(\displaystyle f(0) = f(81)\)

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

Where did this "u" come from?

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

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

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

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

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

\(\displaystyle x = \) :?:

So, after all this you can't do the arithmetic?

What is \(\displaystyle 2^2\)? What is \(\displaystyle 9^2\)? What does the negative power do?

 

[Jason76]

Yes, the function is diffeferentiable and continuous but it is not a polynomial.


Where did this "u" come from?


So, after all this you can't do the arithmetic?

What is \(\displaystyle 2^2\)? What is \(\displaystyle 9^2\)? What does the negative power do?

I would come up with \(\displaystyle x = \dfrac{1}{(\dfrac{2}{9})^{2}}\) which equals \(\displaystyle \dfrac{1}{\dfrac{4}{81}}\)

 

[HallsofIvy]

I would come up with \(\displaystyle x = \dfrac{1}{(\dfrac{2}{9})^{2}}\) which equals \(\displaystyle \dfrac{1}{\dfrac{4}{81}}\)

And? What is a much simpler way of writing \(\displaystyle \dfrac{1}{\dfrac{4}{81}}\)?