Answer
(a) $$g(x)=\frac{1}{1-x} \qquad h(x)=x^2$$
(b) $$g(x)=|x| \qquad h(x)= 5+2x$$
Work Step by Step
(a) By considering the functions $$g(x)=\frac{1}{1-x} \qquad h(x)=x^2$$ we have $$f(x)=(g \circ h)(x)=g(h(x))=\frac{1}{1-x^2}.$$
(b) By considering the functions $$g(x)=|x| \qquad h(x)= 5+2x$$ we have $$f(x)=(g \circ h)(x)=g(h(x))=|5+2x|.$$
