Chapter 0.2, Problem 65E

a.

To determine

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

Explanation of Solution

Given information:

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

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 $g\left(x\right)=\sqrt{x+2}$

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

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

The graph of $f\circ g\left(x\right)=x-1$ is given below.

  

The graph of $g\circ f\left(x\right)=\sqrt{x^2-1}$ 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 $f\left(x\right)=x^2-3,g\left(x\right)=\sqrt{x+2}$ ..

Explanation of Solution

Given information:

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

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 $g\left(x\right)=\sqrt{x+2}$

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

The domain of $f\left(x\right)=x^2-3$ , is any real valued function.

Then domain of $f\circ g\left(x\right)=x-1$ is $x\in R.$

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

The range of the composite function $f\circ g\left(x\right)=x-1$ is $y\in R$ .

The domain of $g\circ f\left(x\right)=\sqrt{x^2-1}$ is $x\in R,x\le -1,x\ge 1$ 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 $y\ge 0$ since it is a root function.