Why Is f'(x)>0 When f''(x)>0 [duplicate]

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Why is $f'(x)>0$ when $f''(x)>0$ on the inequality $\mathrm{e}^x>1+x+\frac{x^2}{2}$ when $x>0$? $f''(x)$ determines the concave of the inequality. When a curve is concave up, it can either go down or up. If it goes down, then how can you say that $f'(x)>0$? I need to prove the equation above is true by doing the above and I don't get the logic in the above steps.

marked as duplicate by quid♦Jul 21 '18 at 17:20

This question was marked as an exact duplicate of an existing question.

• I don't really understand what you are asking and what might be the relationship between the inequality and the derivatives. – Jonas Lenz Jul 21 '18 at 9:35

$f(x):=e^x-1-x-\frac{x^2} {2}$ $f''>0$ so $f'$ is increasing. If $f'(x)=0$, $x=0$. So $f'>0$ when $x>0$.