Proving if crossing line is above function, second derivative is non-negative

Asked Dec 05 '19 at 06:39

Active Dec 05 '19 at 06:39

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Let’s assume $f$ is two times differentiable at all real numbers, and for any $a$ and $b$ that $a<b$, $f$ at $(a, b)$ is under the line drawn between $(a, f(a))$ and $(b, f(b))$.

How could we prove that for any real $x$, the second derivative at $x$ is non negative?

asked Dec 05 '19 at 06:39

Jigsaw

Are you familiar with convex functions? – Kavi Rama Murthy Dec 05 '19 at 06:42
In fact, I wanted to derive that the second derivative is non negative from knowing that the function is convex upward. – Jigsaw Dec 05 '19 at 06:45
Hint: show that the first derivative in increasing: if $x<y$ compare $\frac {f(x+h)-f(x)}x$ with $\frac {f(y+h)-f(y)}x$ for. small positive values of $h$. – Kavi Rama Murthy Dec 05 '19 at 07:16

Proving if crossing line is above function, second derivative is non-negative

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