# Thread: Mean Value Theorem Revisited!

1. ## Mean Value Theorem Revisited!

Let $f(x)=\tan^{-1} x,$ and $0 < a < b.$ By MVT there exists $a < c < b$ such that $f(b)-f(a)=(b-a)f'(c).$ Prove that $\sqrt{ab} < c < \sqrt{\frac{a^2 + b^2}{2}}.$ (obviously $c$ is unique.)

Source: Alex Lupas.

Note: This is just a simple example. The importance of the problem is that it looks at a less discussed side of mean value theorem, i.e. the location of $c.$

2. Well, simple f'=1/(1+x^2) so f'(c)=1/(1+c^2)=(arctg(b)-arctg(a))/(b-a)
c^2=(b-a)/(arctg(b)-arctg(a))-1
c=sqrt(((b-a)/(arctg(b)-arctg(a))-1)
I need to show that (b-a)/(arctg(b)-arctg(a))-1>ab, and I know that the maximum is attained when it equals pi/2, not sure how to continue from here, any hints?

3. Originally Posted by NonCommAlg
Let $f(x)=\tan^{-1} x,$ and $0 < a < b.$ By MVT there exists $a < c < b$ such that $f(b)-f(a)=(b-a)f'(c).$ Prove that $\sqrt{ab} < c < \sqrt{\frac{a^2 + b^2}{2}}.$ (obviously $c$ is unique.)

Source: Alex Lupas.

Note: This is just a simple example. The importance of the problem is that it looks at a less discussed side of mean value theorem, i.e. the location of
$c.$
Claim: $c^2 > ab$

Clearly $\frac{b-a}{1+ab} = \tan\left(\frac{b-a}{1+c^2}\right)$

Since b > a > 0 and $\tan x > x, \forall x > 0$, we get $\frac{b-a}{1+ab} = \tan\left(\frac{b-a}{1+c^2}\right) > \frac{b-a}{1+c^2} \Rightarrow c^2 > ab$

Claim: $c^2 < \frac{a^2 + b^2}{2}$

This part of the proof is not so neat

By a straightforward manipulation, $c^2 = \frac{b - a}{\tan^{-1} b - \tan^{-1} a} - 1 = \frac{\tan x - \tan y}{x - y} - 1$, where $\tan x = b, \tan y = a$.

Hence we have to prove that $c^2 < \frac{\tan^2 x + \tan^2 y}{2}$

The $\tan^{-1}$ function has range from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, we will assume x and y to be in that range.

$b > a > 0 \Rightarrow x > y > 0$ which leads to:

$\frac{\tan x - \tan y}{x - y} = \frac1{\cos^2 x \cos^2 y} \left(\cos x \cos y\frac{\sin (x - y)}{(x - y)} \right) <$ $\frac1{\cos^2 x \cos^2 y} \cos x \cos y < \frac1{\cos^2 x \cos^2 y}\frac{\cos^2 x + \cos^2 y}{2}$ $= \frac{\sec^2 x + \sec^2 y}{2} = \frac{\tan^2 x + \tan^2 y}{2} + 1$

Thus $c^2 = \frac{\tan x - \tan y}{x - y} - 1 < \frac{\tan^2 x + \tan^2 y}{2}$