Welcome to our community

Be a part of something great, join today!

Prove that a function does not have a fixed point

Amer

Active member
Mar 1, 2012
275
it is a question in my book said

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

but i found that this function has a fixed point

[tex] y = 2 + y - \tan ^{-1} y [/tex]

[tex]y = \tan 2 [/tex]

is it right that the question is wrong
 
Last edited by a moderator:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Re: Prove that a function dose not have a fixed point

Hint: $\tan^{-1}(\tan 2)\ne 2$.
 

Amer

Active member
Mar 1, 2012
275

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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, $\tan^{-1}(\tan(2))=2-\pi$.