Chapter 1.3, Problem 32E

a.

To determine

To find:the given function $f\circ g$ and their domain.

Answer to Problem 32E

  $\left(f\circ g\right)\left(x\right)=x^2+3x+2;\text{domain}\in ℝ$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=x-2,g\left(x\right)=x^2+3x+4$ .

Calculation:

As the given function is $f\left(x\right)=x-2,g\left(x\right)=x^2+3x+4$ .

  $\begin{array}{l}\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =\left(x^2+3x+4\right)-2\{\text{as }f\left(x\right)=x-2\}\\ =x^2+3x+2\end{array}$

Note that the domain of the function is a polynomial function thus, $\text{domain}\in ℝ$ .

Hence,the range and domain are $\left(f\circ g\right)\left(x\right)=x^2+3x+2;\text{domain}\in ℝ$

b.

To determine

To find: the given function $g\circ f$ and their domain.

Answer to Problem 32E

  $\left(g\circ f\right)\left(x\right)=x^2-3x+2;\text{domain}\in ℝ$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=x-2,g\left(x\right)=x^2+3x+4$ .

Calculation:

As the given function is $f\left(x\right)=x-2,g\left(x\right)=x^2+3x+4$ .

  $\begin{array}{l}\left(g\circ f\right)\left(x\right)=f\left(g\left(x\right)\right)\\ ={\left(x-2\right)}^2+3\left(x-2\right)+4\{\text{as }g\left(x\right)=x^2+3x+4\}\\ =x^2-6x+4+3x-6+4\\ =x^2-3x+2\end{array}$

Note that the domain of the function is a polynomial function thus, $\text{domain}\in ℝ$ .

Hence, the range and domain are $\left(g\circ f\right)\left(x\right)=x^2-3x+2;\text{domain}\in ℝ$

c.

To determine

To find: the given function $f\circ f$ and their domain.

Answer to Problem 32E

  $\left(f\circ f\right)\left(x\right)=x-4;\text{domain}\in ℝ$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=x-2,g\left(x\right)=x^2+3x+4$ .

Calculation:

As the given function is $f\left(x\right)=x-2,g\left(x\right)=x^2+3x+4$ .

  $\begin{array}{l}\left(f\circ f\right)\left(x\right)=f\left(f\left(x\right)\right)\\ =\left(x-2\right)-2\{\text{as }f\left(x\right)=x-2\}\\ =x-4\end{array}$

Note that the domain of the function is a polynomial function thus, $\text{domain}\in ℝ$ .

Hence, the range and domain are $\left(f\circ f\right)\left(x\right)=x-4;\text{domain}\in ℝ$

d.

To determine

To find: the given function $g\circ g$ and their domain.

Answer to Problem 32E

  $\left(g\circ g\right)\left(x\right)=x^4+6x^3+20x^2+33x+32;\text{domain}\in ℝ$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=x-2,g\left(x\right)=x^2+3x+4$ .

Calculation:

As the given function is $f\left(x\right)=x-2,g\left(x\right)=x^2+3x+4$ .

  $\begin{array}{l}\left(g\circ f\right)\left(x\right)=f\left(g\left(x\right)\right)\\ ={\left(x^2+3x+4\right)}^2+3\left(x^2+3x+4\right)+4\{\text{as }g\left(x\right)=x^2+3x+4\}\\ =x^4+{\left(3x\right)}^2+{\left(4\right)}^2+2x^2\times \left(3x\right)+2\left(3x\right)\left(4\right)+2\left(4\right)x^2+3x^2+9x+12+4\\ =x^4+9x^2+16+6x^3+24x+8x^2+3x^2+9x+12+4\\ =x^4+6x^3+20x^2+33x+32\end{array}$

Note that the domain of the function is a polynomial function thus, $\text{domain}\in ℝ$ .

Hence, the range and domain are $\left(g\circ g\right)\left(x\right)=x^4+6x^3+20x^2+33x+32;\text{domain}\in ℝ$