Prove that a function does not have a fixed point

[Amer]

it is a question in my book said

Prove that the function f(x) = 2 + x - \tan ^{-1} x has the property \mid f'(x)\mid < 1
Prove that f dose not have a fixed point

but i found that this function has a fixed point

y = 2 + y - \tan ^{-1} y

y = \tan 2

is it right that the question is wrong

 

[Evgeny.Makarov]

Re: Prove that a function dose not have a fixed point

Hint: $\tan^{-1}(\tan 2)\ne 2$.

 

[Amer]

Re: Prove that a function dose not have a fixed point

Hint: $\tan^{-1}(\tan 2)\ne 2$.

can you explain more please

 

[Evgeny.Makarov]

Re: Prove that a function dose not have a fixed point

Since tan(x) is periodic, for each y (such as y = tan(2)) there exists an infinite number of x such that tan(x) = y. Therefore, a convention is needed to select a single value for $\tan^{-1}(y)$. By definition, arctangent returns values in the interval $(-\pi/2,\pi/2)$, called principal values. Therefore, $\tan^{-1}(\tan(2))=2-\pi$.