## Calculus: Early Transcendentals (2nd Edition)

$f \circ g=\sqrt{x^3-2}$ $g \circ f=x^{3/2}-2$ $f \circ f=x^{1/4}$ $g \circ g=x^9-6x^6+12x^3-10$
We have: $f(x)=\sqrt{x}$ and $g(x)=x^3-2$. $f \circ g=f(g(x))=\sqrt{x^3-2}$ $g \circ f=g(f(x))=(\sqrt{x})^3-2=x^{3/2}-2$ $f \circ f=f(f(x))=\sqrt{\sqrt{x}}=x^{1/4}$ $g \circ g=g(g(x))=(x^3-2)^3-2=x^9-6x^6+12x^3-10$