# Thread: f(0)=f’(0)=0, f’(x)>0, x>0 -> f(x)>0, x>0

1. ## f(0)=f’(0)=0, f’(x)>0, x>0 -> f(x)>0, x>0

Given f(0)=f’(0)=0, f’(x)>0, x>0
Show: f(x)>0, x>0 using std calculus (no abstract symbolism please).

2. ## Re: f(0)=f’(0)=0, f’(x)>0, x>0 -> f(x)>0, x>0

Question: Is $f(x)$ continuous on $[0,\infty]$ and differentiable on $(0,\infty)$? (This is true for $f(x) = x - \ln(1+x)$).

If so, then we can apply the Mean Value Theorem. Let $a>0$. By assumption, $f(x)$ is continuous on $[0,a]$ and differentiable on $(0,a)$, so by the Mean Value Theorem, there exists $c\in(0,a)$ such that:

$f'(c) = \dfrac{f(a)-f(0)}{a-0} = \dfrac{f(a)}{a}$

Solving for $f(a)$, we have $f(a) = af'(c)$. Since $c>0$, we are given that $f'(c)>0$. The product of two positive real numbers is positive, so $f(a)>0$.