Chapter 1.4, Problem 1E

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

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

Solution

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

The required functions are: $\left(f+g\right)={\left(x+1\right)}^2-2$ and $\left(f-g\right)=-{\left(x-1\right)}^2$ . 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 $f\left(x\right)=2x-1$ ; $g\left(x\right)=x^2$ .

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

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

Since, ${\left(x+1\right)}^2-2$ is a polynomial (algebraic expression), defined for all real numbers. Therefore, the domain of $\left(f+g\right)x$ is $\left(-\infty ,\infty \right)\text{or}ℝ$ .

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

Since, $-{\left(x-1\right)}^2$ is a polynomial (algebraic expression), defined for all real numbers. Therefore, the domain of $\left(f-g\right)x$ is $\left(-\infty ,\infty \right)\text{or}ℝ$ .