## Calculus: Early Transcendentals 8th Edition

$F(x)=h\circ g\circ f$ where $f(x)=x+\sqrt x$ $g(x)=\sqrt x$ $h(x)=\frac{1}{x}$
let: $f(x)=x+\sqrt x$ $g(x)=\sqrt x$ $h(x)=\frac{1}{x}$ $g(f(x))=g(x+\sqrt x)=\sqrt{x+\sqrt x}$ $h(g(f(x)))=h(\sqrt{x+\sqrt x})=\frac{1}{\sqrt{x+\sqrt x}}=F(x)$ $\therefore F(x)=h\circ g\circ f$ where $f(x)=x+\sqrt x$ $g(x)=\sqrt x$ $h(x)=\frac{1}{x}$