SOLUTION 13: Consider the function $ f(x) = \tan x$ on the interval $ [0, \pi/4] $, and assume that $ 0 \le z < w \le \pi/4 $. This function is continuous (well-known) on the closed interval $ [z, w]$. The derivative of $f$ is
$$ f'(x) = \sec^2 x $$
We can now see that $f$ is differentiable on the open interval $ (z, w) $. The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem on the closed interval $ [z, w] $. Then there is some number $c$, $ z < c < w$, so that
$$ f'(c) = \displaystyle{ f(w)-f(z) \over w-z } \ \ \ \ \longrightarrow $$
$$ \sec^2 c = \displaystyle{ \tan w - \tan z \over w-z } \ \ \ \ \longrightarrow $$
$$ \Bigg| \displaystyle{ \tan w - \tan z \over w-z } \Bigg| = | \sec^2 c | \le | \sec^2 (\pi/4) | = | (\sqrt{2})^2| = 2 \ \ \ \ \longrightarrow $$
$\Big($since we assume that $ 0 \le z < w \le \pi/4 \Big)$
$$ \displaystyle{ | \tan w - \tan z | \over | w-z | } \le 2 \ \ \ \ \longrightarrow $$
$$ | \tan w - \tan z | \le 2 | w-z | $$
Click HERE to return to the list of problems.