Chapter 0.2, Problem 66E

a.

To determine

To graph: $f\circ g\left(x\right),g\circ f\left(x\right)$ ,where $f\left(x\right)=\frac{2x-1}{x+3},g\left(x\right)=\frac{3x+1}{2-x}$ and then write a conjecture about domain and range of each function.

Explanation of Solution

Given information:

The given function are $f\left(x\right)=\frac{2x-1}{x+3},g\left(x\right)=\frac{3x+1}{2-x}$

Concept used:

The composition $f\text{of}\text{g}\text{of}\text{x}$ is the composition $f$ with $g$ .The value of $f\circ g\left(x\right)=f\left(g\left(x\right)\right)$ .

Calculations:

Here $g\left(x\right)=\frac{3x+1}{2-x}$

Replace $x$ by $g\left(x\right)=\frac{3x+1}{2-x}$ in $f\left(x\right)$ to get $f\left(g\left(x\right)\right)$ .

  $\begin{array}{c}f\circ g\left(x\right)=f\left(g\left(x\right)\right)\\ f\left(g\left(x\right)\right)=f\left(\frac{3x+1}{2-x}\right)\\ =\frac{2\left(\frac{3x+1}{2-x}\right)-1}{\frac{3x+1}{2-x}+3}\\ =x\\ g\circ f\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(\frac{2x-1}{x+3}\right)\\ =\frac{3\left(\frac{2x-1}{x+3}\right)+1}{2-\frac{2x-1}{x+3}}\\ =\frac{7x}{5}\end{array}$

The graph of $f\circ g\left(x\right)=x$ is given below.

  

The graph of $g\circ f\left(x\right)=\frac{7x}{5}$ is given below

  

Conjecture:

The domain of the composite function $f\circ g\left(x\right)=f\left(g\left(x\right)\right)$ the set of input value for in the domain $g\left(x\right)$ for which $g\left(x\right)$ is the domain of $f\left(x\right)$ in $f\circ g\left(x\right)$ and the range is depend on the newly formed composition function.

Similarly the domain of the composite function $g\circ f\left(x\right)=g\left(f\left(x\right)\right)$ the set of input value for in the domain $f\left(x\right)$ in which $f\left(x\right)$ is the domain of $g\left(x\right)$ in $g\circ f\left(x\right)$ and the range is depend on the newly formed composition function.

b.

To determine

To find : Domain and range of $f\circ g\left(x\right),g\circ f\left(x\right)$ , using conjecture where $f\left(x\right)=\frac{2x-1}{x+3},g\left(x\right)=\frac{3x+1}{2-x}$ ..

Explanation of Solution

Given information:

The given function are $f\left(x\right)=\frac{2x-1}{x+3},g\left(x\right)=\frac{3x+1}{2-x}$

Concept used:

The composition $f\text{of}\text{g}\text{of}\text{x}$ is the composition $f$ with $g$ .The value of $f\circ g\left(x\right)=f\left(g\left(x\right)\right)$ .

The domain of the composite function $f\circ g\left(x\right)=f\left(g\left(x\right)\right)$ the set of input value for in the domain $g\left(x\right)$ for which $g\left(x\right)$ is the domain of $f\left(x\right)$ in $f\circ g\left(x\right)$ and the range is not depend on the inner function .

Similarly the domain of the composite function $g\circ f\left(x\right)=g\left(f\left(x\right)\right)$ the set of input value for in the domain $f\left(x\right)$ in which $f\left(x\right)$ is the domain of $g\left(x\right)$ in $g\circ f\left(x\right)$ and the range is not depend on the inner function.

Calculations:

Here $g\left(x\right)=\frac{3x+1}{2-x}$

The domain of $g\left(x\right)$ is $x\in R,x\ne 2$

The domain of $f\left(x\right)=\frac{2x-1}{x+3}$ , is $x\in R,x\ne -3$ any real valued function.

Then domain of $f\circ g\left(x\right)=x$ is $x\in R.$

Based on the conjecture domain depends on inner and outer function .

The range of the composite function $f\circ g\left(x\right)=x$ is $y\in R$ .

The domain of $g\circ f\left(x\right)=\frac{7x}{5}$ is $x\in R,$ since here inner function is $f\left(x\right)=x^2-3$ so domain of $g\circ f\left(x\right)$ will depend on inner function.

And the range will be also any real value $y\in R$ .