## Calculus 8th Edition

(a) $f(g(x))=-64x^3+48x^2-12x-1$; Domain : $(-\infty, \infty).$ (b) $g(f(x))= 9-4x^3$; Domain : $(-\infty, \infty).$ (c) $f(f(x))= x^9-6x^6+12x^3-10$; Domain : $(-\infty, \infty).$ (d) $g(g(x))= 16x-3$; Domain : $(-\infty, \infty).$
$f(x)=x^3-2$ $g(x) = 1-4x$ (a) $f(g(x))= (1-4x)^3-2 = 1-12x+48x^2-64x^3-2= -64x^3+48x^2-12x-1$ As we have no restrictions, the domain is $(-\infty, \infty).$ (b) $g(f(x))= 1-4(x^3-2) = 1-4x^3+8 = 9-4x^3$ As we have no restrictions, the domain is $(-\infty, \infty).$ (c) $f(f(x))= (x^3-2)^3-2= x^9-6x^5+12x^3-8-2 = x^9-6x^6+12x^3-10$ As we have no restrictions, the domain is $(-\infty, \infty).$ (d) $g(g(x))= 1-4(1-4x) = 1-4+16x=16x-3$ As we have no restrictions, the domain is $(-\infty, \infty).$