Is F(x)=\sqrt{1+x^2} a Contraction Mapping on R?

[flash]

If I take $$F(x)=\sqrt{1+x^2}$$, then the derivative is always less than one so this is a contraction mapping from R to R, right?

But there is no fixed point where $$F(x)=x$$, where the contraction mapping theorem says there should be.

So where have I gone wrong?

Cheers

 

[Dick]

The derivative is less than 1, true. But it approaches 1 as x->infinity. So there is no q<1 such that f'(x)<q. It's NOT a contraction mapping. Look again at the definition of 'contraction mapping'.

 

[flash]

Thanks :-)