# Are these two results applied Mean Value Theorem appropriately?

Are the following two conclusions followed by MVT?

1. $|\tan(x)-\tan(y)| \le |x-y|$, $0\le x,y \le \pi/2$
2. $|2^x-2^y| \ge |x-y|$

For (1), by MVT, $$\frac{\tan(x)-\tan(y)}{x-y}=\sec^2(c) \text{ for some } c \in (0, \pi/2) \tag 3$$
since $\sec^2(c)\ge1$, $c \in (0, \pi/2)$ and it's no harm to put an absolute value on both side of (3), so (1) makes sense.

I think I should get this one correct, though not very confident.

However, for (2), I'm not quite sure:

By MVT, $$\frac{2^x-2^y}{x-y}=2^c\ln2,$$ $c$ is between $x$ and $y$
if $$c\ge0$$ then $$2^c\ge1$$ therefore $$2^x-2^y \ge (\ln2) (x-y)$$ if c<0, then $$0<2^c<1$$ therefore $$2^x-2^y < (\ln2) (x-y) <x-y$$
then I don't know what's next.

And I also have a feeling that my argument for (2) is not complete, when I have the range for $2^c$, I didn't know how to assign $x$ and $y$. for example, if $c<0$, and $c$ is the number between $x$ and $y$, and $x$ can be both negative or one positive and one negative, then I got lost....

Can anyone help me? Thanks a lot.

The second equation is not true in general. Take $y=0$. If $x$ is small, then $2^x - 2^y = e^{x \ln 2} - 1\approx (1 + x \ln 2) - 1 = x \ln 2 < x = x - y$
But the mean value theorem can be used to show when $|2^x - 2^y| \geq |x-y|$ is true. $\frac{2^x - 2^y}{x-y} = (\ln2) 2^c \geq 1$ when $2^c \geq 1/\ln(2)$ or $c \geq \log_2(1/\ln2) \approx 0.53$. So $|2^x - 2^y| \geq |x-y|$ holds when $x, y \geq \log_2(1/\ln2)$
• Your argument is good, just modify it to compute the values of $x$ and $y$ for which the inequality holds. I've updated my answer with the details. – David Goldberg Oct 6 '13 at 7:03