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.

  • $\begingroup$ I don't really understand what you are asking and what might be the relationship between the inequality and the derivatives. $\endgroup$
    – Jonas Lenz
    Commented Jul 21, 2018 at 9:35

1 Answer 1


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


Not the answer you're looking for? Browse other questions tagged .