I have started to use the general form $\frac{f(b)-f(a)}{b-a}$ an substituted the derivative of $\arctan(x)$ on the left side while substituting the intervals to the right side and simplifying all the way down by:
\begin{align*}
\frac{1}{1+x^2} & = \frac{f(0)-f(x)}{0-x}\\
\frac{1}{1+x^2} & = \frac{\arctan(0)-\arctan(x)}{-x}\\
\frac{1}{1+x^2} & = \frac{\arctan(x)}{x}
\end{align*}
What do I do from here because I'm not sure if I've attempted it correctly. How do I find the bounds?
Hint: It is $$\frac{f(b)-f(a)}{b-a}=\frac{\arctan(x)-\arctan(0)}{x-0}=\frac{1}{1+\xi^2}\le 1$$ where $$\xi \in [0,x]$$