MVT Question

J

Jared123

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Have I answered this question correctly? I know I'm missing on some important information but I'm not sure what it is.Q3a.png82436215_292030038855106_3307612962428026880_n.jpg
 
You haven't solved [MATH]2 = \sqrt[3]x[/MATH] correctly. Cube both sides. Then what?
 
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Why do you say the answer is [MATH]\sqrt[3]{3}[/MATH]? And why do you have a [MATH]\pm[/MATH] in your answer?
 
You are missing the point. If f'(x) is NOT differentiable on (-1,8) then YOU CAN NOT USE THE MVT. You can use the mean value theorem ONLY IF f(x) is continuous on [a,b] and differentiable on (a,b). You claim that f(x) is not differentiable on (-1,8) so you should stop and say that the MVT does not guarantee us any c in [-1,8] such that f'(c) = [f(b) - f(a) ]/(b-a).

Please watch the video here as it will show examples of when Rolle's Theorem and the MVT fail.

I appreciate that you stated that f is continuous (but did not say why!!) and that f is not differentiable on (-1,8).

Why is no one else pointing this out!!!
 
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Jomo said:
Why is no one else pointing this out!!!
Click to expand...

The hypotheses of the mean value theorem are not necessary and sufficient for there to exist a point on the open interval satisfying [MATH]\frac {f(b) - f(a)}{b-a}= f'(c)[/MATH]. It is a legitimate question to ask for any such points, which is a different question than to ask to verify the MVT for a particular function/interval. Sometimes such a point exists in spite of the hypotheses not being satisfied.
 
Jomo said:
You are missing the point. If f'(x) is NOT differentiable on (-1,8) then YOU CAN NOT USE THE MVT. You can use the mean value theorem ONLY IF f(x) is continuous on [a,b] and differentiable on (a,b). You claim that f(x) is not differentiable on (-1,8) so you should stop and say that the MVT does not guarantee us any c in [-1,8] such that f'(c) = [f(b) - f(a) ]/(b-a).
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Why is no one else pointing this out!!!
Click to expand...
Because (a) the problem doesn't say to apply the theorem, but just to find such a point, if it exists; and (b) the example is in fact, a counterexample, once you see what the answer is. That's why I challenged both claimed answers.

I would prefer that the problem say not "are satisfied by the mean value theorem", but "satisfy the conclusion of the MVT". A theorem doesn't satisfy a point in the first place!
 
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