Answer
See explanations.
Work Step by Step
a. Given $f(x)=\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}$, we have $f'(x)=\frac{1}{(x+1)^2}\gt0$; thus the function increases on every open interval in its domain ($x\ne-1$).
b. Given $f(x)=x^3+2x$, we have $f'(x)=3x^2+2\gt0$; thus the function increases over $(-\infty,\infty)$. Test end behaviors, $x\to-\infty, f(x)\to-\infty$ and $x\to\infty, f(x)\to\infty$. Also $f'(x)=0$ has no solution, indicating the function has no critical points. Thus, the function has no local maximum or minimum values.
