Find (fg)(x). $f ( x ) = \sqrt { 2 x }$ $g ( x ) = \sqrt { - x + 3 }$
A) $( f g ) ( x ) = x \sqrt { - 2 } + \sqrt { 6 x }$
B) $( f g ) ( x ) = x \sqrt { - 2 + 6 x }$
C) $( f g ) ( x ) = \sqrt { x + 3 }$
D) $( f g ) ( x ) = \sqrt { - 2 x ^ { 2 } + 3 }$
E) $( f g ) ( x ) = \sqrt { - 2 x ^ { 2 } + 6 x }$

Question

Quizplus · Accepted Answer

The composition $(fg)(x) = f(g(x))$ means you substitute $g(x)$ into $f(x)$. So, $f(g(x)) = f(\sqrt{-x + 3}) = \sqrt{2(\sqrt{-x + 3})}$. Squaring the inside gives $-2x + 6$, so the correct expression is $\sqrt{-2x^2 + 6x}$.

Find (Fg)(x) f(x)=2xf ( x ) = \sqrt { 2 x }f(x)=2x​

Find (Fg)(x) $f ( x ) = \sqrt { 2 x }$