Find functions f and g so that f $f \circ g = H$
-$$\mathrm { H } ( \mathrm { x } ) = \sqrt [ 3 ] { \mathrm { x } + 1 }$$
A) $f ( x ) = \sqrt { x } ; g ( x ) = x + 1$
B) $f ( x ) = \sqrt [ 3 ] { x } ; g ( x ) = x + 1$
C) $f ( x ) = x + 1 ; g ( x ) = \sqrt [ 3 ] { x }$
D) $f ( x ) = \sqrt [ 3 ] { x } ; g ( x ) = 1$

Question

Quizplus · Accepted Answer

The correct functions are $f(x) = \sqrt[3]{x}$ and $g(x) = x + 1$, because when you apply $g$ first, you get $x + 1$, and then applying $f$ gives you $\sqrt[3]{x + 1}$, which matches $H(x)$.

Find Functions F and G So That F f∘g=Hf \circ g = Hf∘g=H

Find Functions F and G So That F $f \circ g = H$