Chapter 1, Problem 85RE

a .

To determine

To find: $\left(f+g\right)\left(x\right)$ where $f\left(x\right)=x^2+3$ and $g\left(x\right)=2x-1$

Answer to Problem 85RE

  $\left(f+g\right)\left(x\right)=x^2+2x+2$

Explanation of Solution

Consider,

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

b.

To determine

To find: $\left(f-g\right)\left(x\right)$ where $f\left(x\right)=x^2+3$ and $g\left(x\right)=2x-1$

Answer to Problem 85RE

  $\left(f-g\right)\left(x\right)=x^2-2x+4$

Explanation of Solution

Consider,

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

c .

To determine

To find: $\left(fg\right)\left(x\right)$ where $f\left(x\right)=x^2+3$ and $g\left(x\right)=2x-1$

Answer to Problem 85RE

  $\left(fg\right)\left(x\right)=2x^3-x^2+6x-3$

Explanation of Solution

Consider,

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)\\ =\left(x^2+3\right)\cdot \left(2x-1\right)\\ =2x^3-x^2+6x-3\end{array}$

d.

To determine

To find: $\left(\frac{f}{g}\right)\left(x\right)$ where $f\left(x\right)=x^2+3$ and $g\left(x\right)=2x-1$

Answer to Problem 85RE

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

Domain: All real numbers x except $x=\frac{1}{2}$

Explanation of Solution

Consider,

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

The domain of both $f\left(x\right)$ and $g\left(x\right)$ is the set of real numbers. But the domain of $\left(\frac{f}{g}\right)\left(x\right)$ is the domain common to both $f\left(x\right)$ and $g\left(x\right)$ but with condition $g\left(x\right)\ne 0$ .

Now,

  $\begin{array}{c}g\left(x\right)=0\\ 2x-1=0\\ 2x=1\\ x=\frac{1}{2}\end{array}$

Hence, the domain of $\left(\frac{f}{g}\right)\left(x\right)$ is all real numbers x except $x=\frac{1}{2}$ .