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

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

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

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