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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.2012-01-30

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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 $f(x_1) > g(x_1)$

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    Is it possible to make this result stronger and show that it holds for all x>x_0?2012-01-31