# If $f$ is twice-differentiable at $a$, show that if $f''(a)>0$ then $f(a)+f'(a)(x-a)\leq f(x)$ in a region of $a$.

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 $$a$$.

• The function is differentiable around $a$ but it need not be twice differentiable around $a$ Aug 14 '19 at 12:10

$$f''(a) >0$$ implies that $$f'(x) >f'(a)$$ for $$a for some $$\delta >0$$. [This follows from definition of $$f''(a)$$]. Now $$a 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 $$a$$.