Uniform continuity of $\frac{1}{x^4+1}$ using mean value theorem

Question

Let $g: \mathbb R \to \mathbb R$ and $g(x)=\frac{1}{x^4+1}$.

I am required to prove that $g$ is uniformly continuous on $\mathbb R$ using the mean value theorem.

My path: Let $x,y \in \mathbb R, x<y$ and using the MVT we find an $s\in ]x,y[$ so that $|f'(s)|=|\frac{f(y)-f(x)}{y-x}|$. This then turns into:

$$\left|-\frac{4s^3}{(1+s^4)^2}\right||y-x|=\left|\frac{x^4-y^4}{(y^4+1)(x^4+1)}\right|$$

and this is where I am stuck, I cannot seem to find an appropriate inequality that leads into $|y-x|<\delta$. Any help would be greatly appreciated.

Answer 1

Two things:

1. $f'(s)=-\dfrac{4s^3}{(s^4+1)^2}$.
2. It turns out that $(\forall x\in\mathbb{R}):\bigl|f'(x)\bigr|\leqslant\dfrac{5\sqrt[4]{135}}{16}$ (prove it). Therefore, for each $\varepsilon>0$ you can take $\delta=\dfrac{16\varepsilon}{5\sqrt[4]{135}}$.

Answer 2

For every pair $x,y$, there is a point $s$ between them such that $$ g(x) - g(y) = g'(s)(x-y) \implies |g(x)-g(y)| = |g'(s)| |x-y| $$ If you can show that $|g'(t)| \leq M$ for all $t \in \mathbb{R}$, for some $M$, then $\delta = \frac{\epsilon}{M}$ will do.