# Mean value theorem! (1 Viewer)

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#### 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

Homework Helper
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

Staff Emeritus
Gold Member
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 $f(c) = x$.

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#### NateTG

Homework Helper
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.

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