# How to prove $f(x)>g(x)$ by showing $f'(x)>g'(x)$ for some $x>x_0$

Is it possible to show that some function $f(x)$ is larger than $g(x)$ if I can show that its derivative is larger for some $x>x_0$?

As an example I can think of $f(x)=(1+x)^n , \ g(x)=1+nx+ \frac{n(n-1)x^2}{2}$, which derivatives are $n(1+x)^{n-1}$ and $n+n(n-1)x$ and the use Bernoulli inequality to show that $f'(x)>g'(x)$ for $x>0$.

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No. Try $f(x)=1$ (derivative zero) and $f(x)=-\frac 1 x$ (derivative positive for positive $x$) . So you are missing a condition you need. –  Mark Bennet Jan 30 '12 at 5:30

Just showing that $f'(x) > g'(x)$ for some $x > x_0$ is not enough. If $f(x_0) \ge g(x_0)$ and $f'(x) > g'(x)$ for all $x$ in the interval $(x_0, x_1)$ (and $f$ and $g$ are continuous at $x_0$ and $x_1$, and $x_0 < x_1$), then $f(x_1) > g(x_1)$
Mean Value Theorem. Let $x>x_0$. By the MVT applied to $f-g$ on $[x_0,x]$, there exists a $c \in (x_0,x)$ so that $\frac{f(x)-g(x)- [f(x_0)-g(x_0)]}{x-x_0}=f'(c)-g'(c)$. You should get it imediatelly from there, WHY? –  N. S. Jan 30 '12 at 6:10
Is it possible to make this result stronger and show that it holds for all $x>x_0$? –  user19821 Jan 31 '12 at 6:01