# Mean Value Theorem

• Mar 26th 2009, 04:51 PM
storage1k
Mean Value Theorem
Well, I honestly have no clue how to do this and have doing it based on examples of normal non-trigonometric functions...
I've gotten up to the point of merely differentiating f(x) before not knowing where to go....

Prove using the Mean Value Theorem:

If x>0, then arctan(x) > 1/ (1+x^2)

• Mar 26th 2009, 06:13 PM
storage1k
Hmm, is this correct?

f'(c)(b-a)= ∫[a,b] f(t) dt

Let f(t) = 1/(1+t^2)

b=x and a=0

f'(c)(x-0)= ∫[0,x] 1/(1+t^2) dt = arctan(x)

For 0<c<x,

arctan(x) = x/(1+c^2)

since c<x and since numbers are positive,

x/(1+c^2) > x/(1+x^2) and thus arctan(x)>x/(1+x^2)
• Mar 27th 2009, 03:25 AM
mr fantastic
Quote:

Originally Posted by storage1k
Well, I honestly have no clue how to do this and have doing it based on examples of normal non-trigonometric functions...
I've gotten up to the point of merely differentiating f(x) before not knowing where to go....

Prove using the Mean Value Theorem:

If x>0, then arctan(x) > 1/ (1+x^2)

Here's an outline of how I'd do it:

Consider $f(x) = \arctan x$ and note that $f'(x) = \frac{1}{1 + x^2}$.

From the Mean Value Theorem: $f'(c) = \frac{f(b) - f(a)}{b - a}$, where $a < c < b$.

Apply the MVT to the interval $[0, x]$ and let $c = v$:

$f'(v) = \frac{f(x) - f(0)}{x - 0}$

$\Rightarrow \frac{1}{1 + v^2} = \frac{\arctan x - \arctan 0}{x - 0} = \frac{\arctan x}{x}$

$\Rightarrow \frac{x}{1 + v^2} = \arctan x$.

Now note that $\frac{x}{1 + v^2} > \frac{x}{1 + x^2}$ since $0 < v < x$.

Therefore $\arctan x = \frac{x}{1 + v^2} > \frac{x}{1 + x^2}$.

You should now continue and do the following:

1. Prove $\frac{x}{1 + x^2} < \arctan x < x$.

2. Use part 1. to show that if $0 < \theta < \frac{\pi}{2}$ then $\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}$.

3. Use part 2. to prove that $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$.