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$f(x)=4-(6-x)^{\frac23}$

Well, $f(5)=3, f(7)=3$, so, $\frac{f(7)-f(5)}{7-5}=0$

And $f'(x)= \frac23 (6-x)^{-\frac13}$

So, by MVT, $\exists c\in (5,7)$ such that $\frac{f(7)-f(5)}{7-5}=f'(c)$,

i.e. $\frac23 (6-c)^{-\frac13}=0$, which doesn't yield any solution.

Where did I go wrong?

Note: One possibility that comes to my mind is that the conditions of MVT are not satisfied here, but if so, I cannot identify that. Please help.

Diya
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1 Answers1

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The function is continuous in [5,7], but it is not differentiable for x=6, which belongs to that interval, so you cannot apply the Mean value theorem

Mosk
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