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.

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- Jan 8th 2012, 12:23 PMgrandunificationDerivative 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 AMOpalgRe: Derivative question
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 $\displaystyle x>x_0.$