Q. 44 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 29 – 44, for the given functions f and g, find: $\left(a\right) f\circ g \left(b\right) g\circ f \left(c\right) f\circ f \left(d\right) g\circ g$

$f\left(x\right)=\frac{2x-1}{x-2},g\left(x\right)=\frac{x+4}{2x-5}$

Step-by-Step Solution

Answer

(a) $f\circ g=\frac{13}{-3x+4}$and domain is $R-\left\{\frac{4}{3}\right\}$

(b) $g\circ f=\frac{6x-9}{-x+8}$and domain is $R-\left\{8\right\}$

(c) $f\circ f=x$ and domain is the set of all real number.

(d) $g\circ g=\frac{9x-16}{-8x+33}$and domain is $R-\left\{\frac{33}{8}\right\}$

Step 1. Given Information

The given functions are $f\left(x\right)=\frac{2x-1}{x-2},g\left(x\right)=\frac{x+4}{2x-5}$

Part (a) Step 1. Finding f ∘ g

$$\begin{gathered} f\circ g=f\left(g\left(x\right)\right)=f\left(\frac{x+4}{2x-5}\right) \\ f\circ g=\frac{2\left(\frac{x+4}{2x-5}\right)-1}{\frac{x+4}{2x-5}-2} \\ f\circ g=\frac{{\displaystyle \frac{2x+8-2x+5}{2x-5}}}{{\displaystyle \frac{x+4-4x+10}{2x-5}}} \\ f\circ g=\frac{13}{-3x+14} \end{gathered}$$

Part (a) Step 2. Finding domain of f ∘ g

Put $$\begin{gathered} -3x+14=0 \\ -3x=-14 \\ x=\frac{14}{3} \end{gathered}$$

The domain is the set of all real number except $\frac{14}{3}.$

Part (b) Step 1. Finding g ∘ f

$$\begin{gathered} g\circ f=g\left(f\left(x\right)\right)=g\left(\frac{2x-1}{x-2}\right) \\ g\circ f=\frac{{\displaystyle \frac{2x-1}{x-2}+4}}{2\left(\frac{2x-1}{x-2}\right)-5} \\ g\circ f=\frac{{\displaystyle \frac{2x-1+4x-8}{x-2}}}{{\displaystyle \frac{4x-2-5x+10}{x-2}}} \\ g\circ f=\frac{6x-9}{-x+8} \end{gathered}$$

Part (b) Step 2. Finding domain of g ∘ f

Put $$\begin{gathered} -x+8=0 \\ -x=-8 \\ x=8 \end{gathered}$$

The domain is the set of all real number except

Part (c) Step 1. Finding f ∘ f

$$\begin{gathered} f\circ f=f\left(f\left(x\right)\right)=f\left(\frac{2x-1}{x-2}\right) \\ f\circ f=\frac{2.{\displaystyle \frac{2x-1}{x-2}}-1}{{\displaystyle \frac{2x-1}{x-2}}-2} \\ f\circ f=\frac{4x-2-x+2}{2x-1-2x+4}=\frac{3x}{3}=x \end{gathered}$$

Part (d) Step 2. Finding domain of f ∘ f

We get $f\circ f=x$

This is an identity function. The domain of identity function is the set of  Real number.

Part (d) Step 1. Finding g ∘ g

$$\begin{gathered} g\circ g=g\left(g\left(x\right)\right)=g\left(\frac{x+4}{2x-5}\right) \\ g\circ g=\frac{{\displaystyle \frac{x+4}{2x-5}}+4}{2\left(\frac{x+4}{2x-5}\right)-5}=\frac{{\displaystyle \frac{x+4+8x-20}{2x-5}}}{{\displaystyle \frac{2x+8-10x+25}{2x-5}}} \\ g\circ g=\frac{9x-16}{-8x+33} \end{gathered}$$

Part (d) Step 2. Finding domain

Put $$\begin{gathered} -8x+33=0 \\ -8x=-33 \\ x=\frac{33}{8} \end{gathered}$$