"Does this function satisfy the hypotheses of the Mean Value Theorem, or does it not? (My book is telling me it does, but I can't understand how. The way I calculate the derivative, I end up with an indeterminate form when x = 0 is applied, and 0 is in the selected domain. Question below.

Question

agent0smith · Answer

You never posted the question...

agent0smith · Answer

MVT, where a < c < b
$$f'(c) = \frac{ f(b) - f(a) }{ b-a }$$

mendicant_bias · Answer

Give me a minute, bunch of stuff happened. Will post later.

mendicant_bias · Answer

Okay. Question is $$f(x) = \sqrt{x(1-x)}$$Derivative is $$\frac{ 1 - 2x }{ 2\sqrt{x(1-x)} }$$The closed interval for x in this situation is [0, 1]. I'm relatively sure but not 100% that the reason that it is still applicable by Rolle's Theorem is that only the interior of the interval needs to be differentiable? Because clearly, x = 0 would result in an indeterminate form.

agent0smith · Answer

http://tutorial.math.lamar.edu/Classes/CalcI/MeanValueTheorem.aspx Suppose is a function that satisfies all of the following. is continuous on the closed interval [a,b]. is differentiable on the open interval (a,b). Don't these both apply here. The function is continuous on [0,1] and differentiable on (0,1)

agent0smith · Answer

http://www.sosmath.com/calculus/diff/der11/der11.html $$f'(c) = \frac{ f(b) - f(a) }{ b-a } $$ it looks like we have f(b)=f(a) =0, $$f'(c) = \frac{ 0 - 0 }{ 1-0 }$$

agent0smith · Answer

The function is not differentiable at x=1 either btw, both 0 and 1 give a denominator of zero

I think your derivative looks off also
$$\large \frac{ d }{ dx } f(x) = (x^2-x)^{0.5} = 0.5(x^2-x)^{-0.5}(2x-1)$$ $$\large = \frac{ 2x -1  }{ 2 \sqrt{ x(x-1)} }  $$

mendicant_bias · Answer

Oh, wait, no, actually I was right. There wasn't a typo. Now I really don't have any clue what you're talking about. I checked with Wolfram Alpha and another derivative calculator. http://www.wolframalpha.com/input/?i=%28x%281-x%29%29^%281%2F2%29 http://www.derivative-calculator.net/#

mendicant_bias · Answer

(And yes, I understand the problem. I totally just forgot that it didn't need to be differentiable at its endpoints.) Just curious about where the x^2 - x came from.

agent0smith · Answer

My bad, I must've read the original problem as sqrtx(x-1) not sqrtx(1-x). Your derivative is right.

agent0smith · Answer

But yes, if you plot the graph, you can easily see the horizontal tangent line between 0 and 1: https://www.google.com/search?q=sqrt(x(1-x))&aq=f&oq=sqrt(x(1&aqs=chrome.0.59j57j0l2.4001&sourceid=chrome&ie=UTF-8