Apply Mean Value Theorem to Show arctan x - x = 0 at x = 0

[sonya]

oks...im having problems applying the mean value theorem...i understand the concept behind it but whenever i try 2 do a question i have no idea wat 2 do...heres 1 question I've been trying:

Showing all your work, apply the Mean Value Theorem to show
that the function arctan x - x is equal to zero only at x = 0 .

where do i start? thx.

 

[HallsofIvy]

The mean value theorem says that if f is differentiable on (a,b) then for some c between a and b, f'(c)= (f(b)- f(a))/(b-a).

Certainly tan(x) is differentiable for x between -pi/2 and pi/2 so also is tan(x)- x.

Suppose tan(b)- b= 0 for some b (between -pi/2 and pi/2). Apply the mean value theorem between 0 and b.
There must be a point c between 0 and b so that f'(c)= (f(b)- f(a))/(b-a) = 0. But for f(x)= tan(x)- x, f'(x)= sec²(x)- 1.
f'(x)= 0 means sec²(x)= 1 or sec(x)= +/- 1. That only happens at 0 and pi/2 NOT between 0 and 1.

 

[NateTG]

Huh. That's not the mean value theorem that I'm familiar with. (In fact, it's not true unless the derivative is contiuous.)

The version I'm familiar with is more like:

If $$f$$ is continuous on $$[a,b]$$ then there is some point $$c; a \leq c \leq b$$ so that $$f(c)=\frac{f(a)+f(b)}{2}$$

Now, you know that
$$f(x)=\arctan{x}-x$$
is continuous.
If you can show that it is always less than zero for $$x>0$$ and alwauys greather than zero for $$x<0$$, and that there must be some point where it is zero (mean value theorem might be usefull here), then you're done.

 

[Hurkyl]

I think you're thinking of the intermediate value theorem:

If $f(x)$ is continuous on the interval $[a, b]$, then for any $x$ between $f(a)$ and $f(b)$, there exists a $c \in (a, b)$ such that $f(c) = x$.

 

[NateTG]

I guess that's what getting too much higher math does to my brain. ...
Not to mention that it's easy prove that |arctan x| < |x| for (x &neq; 0) without resorting to calculus.