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Prove that there is at least one point $\varepsilon$, $a\lt \varepsilon \lt b$, for which $f''(\varepsilon) \lt 0$.

My approach : By Rolle's Theorem , $ \exists \alpha \in (a,b)$ such that $f'(\alpha) = 0$.

And as $f(x)$ is differentiable in $[a,b]$ it is also differentiable in $[a,c]$ then by Lagrange Mean value theorem, $\exists \,\beta \in (a,c)$ such that $f'(\beta) = \frac{f(c)-f(a)}{c-a} \gt 0$ since $f(c)>0$, $f(a) = 0$ and $c>a$.

Now assuming $ \alpha \gt \beta$ , and since $f(x)$ is twice differentiable, applying LMVT on $f'(x)$ in the interval $(\beta,\alpha)$, we get that there exists an $\varepsilon \in (\beta,\alpha)$ such that $f''(\varepsilon) = \frac{f'(\alpha)-f'(\beta)}{\alpha - \beta} \lt 0$.

I have assumed $\alpha \gt \beta$, is there a concrete way to prove without assuming? Thank You.

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    $\begingroup$ Don't say "$\exists\; a \;\alpha$" and "$\exists\; a \;\beta$". It's just "$\exists\;\alpha$" and "$\exists\;\beta$". (Also, it's probably not a good idea to use $\varepsilon$ as a generic variable name in the context of Analysis.) $\endgroup$ – TonyK Jun 17 '16 at 16:32
  • $\begingroup$ Furthermore, don't use so many logic symbols, your reader is a person, not a computer. We like words. $\endgroup$ – YoTengoUnLCD Jun 17 '16 at 16:43
  • $\begingroup$ @TonyK Agreed. Here, $\xi$ (xi) is more common than $\varepsilon$. $\endgroup$ – MCT Jun 17 '16 at 16:58
  • $\begingroup$ I didnt know how to write the above symbol. will keep in mind when i write my approach and use less symbols. Thank You $\endgroup$ – Rohith Penumala Jun 17 '16 at 17:22
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As you suspect, your argument is flawed. Here is a simpler way:

$f(c) > 0$, so by the Mean Value Theorem, there exist $\alpha \in (a,c)$ and $\beta \in (c,b)$ such that $f'(\alpha) > 0$ and $f'(\beta) < 0$.

Now use the Mean Value Theorem on the function $f'$ and the interval $(\alpha,\beta)$.

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  • $\begingroup$ Thank You so much for clarifying @TonyK $\endgroup$ – Rohith Penumala Jun 17 '16 at 17:19
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By contradiction, assume $\forall x \in ]a,b[, f''(x) \geq 0$.

Then $f'$ is increasing on $]a,b[$ and thus increasing on $[a,b]$ by continuity.

Because $f(c)>0$ for all $c \in ]a,b[$, we must have $f'(a) \geq 0$.

By Rolle, there is $\alpha \in ]a,b[$ such that $f'(\alpha)=0$.

Let $x \in [a,\alpha]$.

Then $0=f'(\alpha) \geq f'(x) \geq f'(a) \geq 0$ so $f'(x)=0$.

Hence $f'=0$ on $[a,\alpha]$ so $f$ is constant on $[a,\alpha]$ which contradicts the fact that $f(c)>0$ for all $c \in ]a,b[$.

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  • $\begingroup$ Thank you very much for an alternative solution $\endgroup$ – Rohith Penumala Jun 17 '16 at 17:20

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