Q 19. - Precalculus Enhanced with Graphing Utilities

Question

Find for the given functions $f,g$ such that

$f\left(x\right)=\frac{3}{x+1} , g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$.

Find : (a) $\left(f\circ g\right)\left(4\right)$ (b) $\left(g\circ f\right)\left(2\right)$ (c) $\left(f\circ f\right)\left(1\right)$ (d) $\left(g\circ g\right)\left(0\right)$.

Step-by-Step Solution

Answer

The values are : (a) $\frac{3}{2^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1}$ (b) $1$ (c) $\frac{6}{5}$ (d) 0.

Part (a) Step 1. Given Information

The given functions are : 

$f\left(x\right)=\frac{3}{x+1} , g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$.

Part (a) Step 2. The Evaluation of a functional value.

The functional value are :

(a) $\left(f\circ g\right)\left(4\right)=f\left(g\left(4\right)\right)=f\left(2^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)=\frac{3}{2^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+1}$ .

(b) $\left(g\circ f\right)\left(2\right)=g\left(f\left(2\right)\right)=g\left(1\right)=1$.

(c) $\left(f\circ f\right)\left(1\right)=f\left(f\left(1\right)\right)=f\left(\frac{3}{2}\right)=\frac{6}{5}$.

(d) $\left(g\circ g\right)\left(0\right)=g\left(g\left(0\right)\right)=g\left(0\right)=0$.

Part (b) Step 1. Given Information.

The given function be

$f\left(x\right)=\frac{3}{x+1} , g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$.

Part (b) Step 2. Evaluation of a functional value.

The functional value be

$\left(g\circ f\right)\left(2\right)=g\left(f\left(2\right)\right)=g\left(1\right)=1$.

Part (c) Step 1. Given Information.

The given function be

$f\left(x\right)=\frac{3}{x+1} , g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$.

Part (c) Step 2. Evaluation of a functional value.

The functional value be

$\left(f\circ f\right)\left(1\right)=f\left(f\left(1\right)\right)=f\left(\frac{3}{2}\right)=\frac{6}{5}$.

Part (d) Step 1. Given Information.

The given function be

$f\left(x\right)=\frac{3}{x+1} , g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$.

Part (d) Step 2. Evaluation of a functional value.

The functional value be

$\left(g\circ g\right)\left(0\right)=g\left(g\left(0\right)\right)=g\left(0\right)=0$.