Show that $f(x) =\frac{1}{1+x^2}$ is Lipschitz continuous on $\mathbb{R}$?

Question

Show that $$f(x) = \frac{1}{1+x^2}$$ is Lipschitz continuous on $\mathbb{R}$?

I know $0<f(x)<1$ but $x$ and $y$ are any number in $\mathbb{R}$. How do you find a constant $c$ such that $$|f(x) - f(y)| < c |x - y| \ ?$$

All answers there are about Lipschitz continuity. – Jul 09 '15 at 00:56 — , Jul 09 '15 at 00:56

score -1 · Answer 1 · answered Jul 09 '15 at 00:57

-1

hint: $|f'(x)| = \left|\dfrac{-2x}{(1+x^2)^2}\right|\leq \dfrac{1}{1+x^2} \leq 1$

answered Jul 09 '15 at 00:57

DeepSea

1 Answers1