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We have two differentiable functions from $\mathbb{R} \rightarrow \mathbb{R}$; suppose that $f'(x) \leq g'(x)$ for every real number $x$ and $f(0) = g(0)$; then show that $f(x) \leq g(x)$ when $x \geq 0$ and $f(x) \geq g(x)$ when $x \leq 0$

Would it be correct to use the extended mean value theorem to prove this? For instance to write there is a point $x \in (0, b)$ at which $$\frac{f(b) - f(0)}{g(b) - g(0)} = \frac{f'(x)}{g'(x)} \leq 1$$ The issue is that I can't guarantee the sign of $g'(x)$, so I'm not even sure if this works.

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    $\begingroup$ Did you forget to include a hypothesis that $f(0)=g(0)$? $\endgroup$ – Michael Hardy Mar 31 '12 at 23:09
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    $\begingroup$ I think you should defining new function $h:\mathbb{R}\to\mathbb{R}$ as $h(x)=f(x)-g(x)$ ... $\endgroup$ – Salech Rubenstein Mar 31 '12 at 23:09
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Just working with $g-f$ and applying the simple mean value theorem of first-year calculus will do it, provided you have a hypothesis that $f(0)=g(0)$. Without that last assumption, the conclusion you're trying to get is not true, and it's not hard to find counterexamples.

So you get $$\frac{(g-f)(x) - (g-f)(0)}{x-0} = (g-f)'(c) \ge 0$$ for some $c$ between $0$ and $x$. Multiply both sides by $x$. If $x>0$, then the inequality still says "$\ge$"; if $x<0$, it says "$\le$". And use the fact that $(g-f)(0)=0$.

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