## Precalculus (6th Edition) Blitzer

(a) Consider the function \begin{align} & \left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right) \\ & =f\left( x+1 \right) \\ & =\sqrt{x+1} \end{align} Therefore, the value of $\left( f\circ g \right)\left( x \right)$ for the functions $f\left( x \right)=\sqrt{x}$ and $g\left( x \right)=x+1$ is $\sqrt{x+1}$. (b) Consider the function \begin{align} & \left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right) \\ & =g\left( \sqrt{x} \right) \\ & =\sqrt{x}+1 \end{align} Therefore, the value of $\left( g\circ f \right)\left( x \right)$ for the functions $f\left( x \right)=\sqrt{x}$ and $g\left( x \right)=x+1$ is $\sqrt{x}+1$. (c) From part (a), $\left( f\circ g \right)\left( x \right)=\sqrt{x+1}$. Therefore, \begin{align} & \left( f\circ g \right)\left( 3 \right)=\sqrt{3+1} \\ & =\sqrt{4} \\ & =\pm 2 \end{align} Therefore, the value of $\left( f\circ g \right)\left( 3 \right)$ for the functions $f\left( x \right)=\sqrt{x}$ and $g\left( x \right)=x+1$ is $\pm 2$.