# Derivative question

• Jan 8th 2012, 12:23 PM
grandunification
Derivative question
Let f be a function from R to R, and f ' (x) > f (x) for all x. Suppose f(x_0) =0. Show that f (x) > 0 for all x > x_0.

Thanks!
• Jan 9th 2012, 07:48 AM
Opalg
Re: Derivative question
Quote:

Originally Posted by grandunification
Let f be a function from R to R, and f ' (x) > f (x) for all x. Suppose f(x_0) =0. Show that f (x) > 0 for all x > x_0.

Since \$\displaystyle f'(x_0) > f(x_0) = 0\$, it follows that f is increasing at \$\displaystyle x_0.\$ Therefore there is some interval to the right of \$\displaystyle x_0\$ in which f is positive. Suppose by way of getting a contradiction that f is not positive for all \$\displaystyle x > x_0.\$ Then (by continuity, intermediate value theorem and suchlike properties) there must be a smallest point \$\displaystyle x_1>x_0\$ at which \$\displaystyle f(x) = 0.\$ Thus \$\displaystyle f(x)>0\$ whenever \$\displaystyle x_0<x<x_1.\$ But since \$\displaystyle f(x_0) = f(x_1) = 0\$ it follows from Rolle's theorem that there is a point \$\displaystyle x_2\$ with \$\displaystyle x_0<x_2<x_1\$ such that \$\displaystyle f'(x_2)=0.\$ Then \$\displaystyle f'(x_2)<f(x_2)\$, contradicting the fact that \$\displaystyle f'(x)>f(x)\$ for all \$\displaystyle x>x_0.\$