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Given some function $f: I \subseteq\mathbb R \rightarrow \mathbb R$, Which is differentiable twice at some point $a\in I$.

Prove:

If $f''(a)>0$ then, $f(a)+f'(a)(x-a)\leq f(x)$ in a region of $a$.

I think we can't use the Lagrange remainder for taylor polynomials, because we can't assume that the function is differentiable around $a$.

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    $\begingroup$ The function is differentiable around $a$ but it need not be twice differentiable around $a$ $\endgroup$ – Kavi Rama Murthy Aug 14 at 12:10
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$f''(a) >0$ implies that $f'(x) >f'(a)$ for $a <x <a+\delta$ for some $\delta >0$. [This follows from definition of $f''(a)$]. Now $a <x <a+\delta$ implies $f(x)-f(a)=(x-a)f'(t)$ for some $t \in (a,x)$ so $f(x)-f(a)\geq (x-a)f'(a)$. Similar argument works to the left of $a$.

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