Solutions

a) False. As a counterexample, let $f(x)=x^2$ and $g(x)=h(x)=x$. Then we have that

$$f(g(x)+h(x))=f(2x)=4x^2\ne 2x^2=f(g(x))+f(h(x)).$$

b) True. We have that

$$\begin{align*} ((g+h)\circ f)(x) &= (g+h)(f(x)) \\ &= g(f(x))+h(f(x)) \\ &= (g\circ f)(x)+(h\circ f)(x) \\ &= ((g\circ f)+(h\circ f))(x). \end{align*}$$

c) True. Let $h(x)=\frac{1}{f(x)}$. We have that

$$\begin{align*} \left(\frac{1}{f}\circ g\right)(x) &= (h\circ g)(x) \\ &= h(g(x)) \\ &= \frac{1}{f(g(x))} \\ &= \frac{1}{(f\circ g)(x)}. \end{align*}$$

d) False. Let $f(x)=g(x)=2$. Then, for $x=0$, we have

$$\frac{1}{f(g(0))}=\frac{1}{2}\ne 2=f\left(\frac{1}{2}\right)=f\left(\frac{1}{g(0)}\right).$$