Chapter 2.7, Problem 51E

Solution

a.

To find: The combined form of the fractions in the function.

The combined form is $f\left(x\right)=\frac{x^2+2x}{x^2+2x}$ .

Given:

The function is $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$ .

Calculation:

The function is $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$ .

Take the L.C.M of the denominator $x$ , $x+2$ and $x^2+2x$ and the L.C.M is $x^2+2x$ .

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

b.

To find: The domain of the function $f\left(x\right)=\frac{x^2+2x}{x^2+2x}$ .

The domain of $f\left(x\right)$ is $\left\{x\in \left(-\infty ,\infty \right),x\ne -2,0\right\}$ .

Given:

The function is $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$ .

Calculation:

The domain of any real function is the set of all real numbers except those values of $x$ for which the function becomes undefined.

From the function it is clear that the function becomes undefined at $x=0$ , $-2$ .

Conclusion:

The domain of the function is $\left\{x\in \left(-\infty ,\infty \right),x\ne -2,0\right\}$

c.

To find: The piece-wise function of the function $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$ .

The piece-wise defined function is $f\left(x\right)=\left\{\begin{array}{l}1,x\ne 0,-2\\ \text{undefined, }x=0,-2\end{array}\right\}$ .

Given:

The function is $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$ .

Calculation:

The function is $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$ .

The function can also be written as shown below.

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

The function is defined and the value of the function would be $1$ , undefined at $x=0$ , $-2$ .

Conclusion:

The piece-wise function is $f\left(x\right)=\left\{\begin{array}{l}1,x\ne 0,-2\\ \text{undefined, }x=0,-2\end{array}\right\}$

d.

To find: The graph of the function $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$

The graph of the function is as shown below.

  

Given:

The function is $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$ .

Calculation:

Press the $Y=$ button and enter the functions $f\left(x\right)=\frac{x-3}{x}+\frac{3}{x+2}+\frac{6}{x^2+2x}$ .

  

Conclusion:

The above graph shows that the domain of the function is $\left\{x\in \left(-\infty ,\infty \right),x\ne -2,0\right\}$ and piece-wise defined function is $f\left(x\right)=\left\{\begin{array}{l}1,x\ne 0,-2\\ \text{undefined, }x=0,-2\end{array}\right\}$ .