Answer
See explanations.
Work Step by Step
a). Step 1. We work on the left inequality first: to show that $\frac{x}{1+x^2}\geq -\frac{1}{2}$, we need to have $2x\geq –(1+x^2)$ or $x^2+2x+1\geq 0$. This is always true because $x^2+2x+1=(x+1)^2\geq 0$.
Step 2. For the right half of the inequality, we need to show that $\frac{x}{1+x^2}\leq \frac{1}{2}$. This is equivalent to $2x\leq (1+x^2)$ or $x^2-2x+1\geq 0$. Again, this is always true because $x^2-2x+1=(x-1)^2\geq 0$.
b). Given $f’(x)= \frac{x}{1+x^2}$, based on the results in part (a), we have $-\frac{1}{2}\leq f’(x)\leq \frac{1}{2}$ or $|f’(x)|\leq\frac{1}{2}$ for all the $x$ in the interval. Based on the Intermediate Value Theorem, there exists a value $c$ in the interval $[a,b]$ (or $[b,a]$) such that $f’(c)=\frac{f(b)-f(a)}{b-a}$. Since $|f’(c)|\leq\frac{1}{2}$, we have $|f’(c)|=|\frac{f(b)-f(a)}{b-a}|\leq\frac{1}{2}$ or $|f(b)-f(a)|\leq \frac{1}{2}|b-a|$.
