Chapter 1.4, Problem 2E

In Exercises 1−4 find formulas for the functions $f+g,f-g$ , and $fg$ . Give the domain of each.

2. $f\left(x\right)={\left(x-1\right)}^2;g\left(x\right)=3-x$

Solution

To find: The formula and domain for functions $f+g$ $f+g$ .

The required functions are: $\left(f+g\right)=x^2-3x+4$ and $\left(f-g\right)=x^2-x-2$ . The domain of each function $\left(f+g\right)$ and $\left(f-g\right)$ is set of real numbers $\left(-\infty ,\infty \right)\text{or}ℝ$ .

Given data: The functions $f\left(x\right)$ and $g\left(x\right)$ are $f\left(x\right)={\left(x-1\right)}^2$ and $g\left(x\right)=3-x$ .

Method/Formula used:

The sum and difference of the two functions $f\left(x\right)$ and $g\left(x\right)$ is also a function given by $f\left(x\right)+g\left(x\right)=\left(f+g\right)x$ , and $f\left(x\right)-g\left(x\right)=\left(f-g\right)x$ . The domain of resulting function $\left(f+g\right)x$ and $\left(f-g\right)x$ are the values of values of x that define $\left(f+g\right)x$ and $\left(f-g\right)x$ .

Calculation:

The given functions are $f\left(x\right)={\left(x-1\right)}^2$ ; $g\left(x\right)=3-x$ . Therefore, using the definition of $f\left(x\right)+g\left(x\right)$ and $f\left(x\right)-g\left(x\right)$ are:

  $\begin{array}{c}\left(f+g\right)x=f\left(x\right)+g\left(x\right)\\ ={\left(x-1\right)}^2+\left(3-x\right)\\ =x^2-2x+1+3-x\\ =x^2-3x+4\end{array}$

Since, $x^2-3x+4$ is a polynomial function (algebraic expression), defined for all real numbers. Therefore, the domain of $\left(f+g\right)x$ is $\left(-\infty ,\infty \right)\text{or}ℝ$ .

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

Since, $x^2-x-2$ is a polynomial (algebraic expression), defined for all real numbers. Therefore, the domain of $\left(f-g\right)x$ is $\left(-\infty ,\infty \right)\text{or}ℝ$ .