Question

evaluate the function $f(x) = \frac{x}{|x|}$ at the given values of the independent variable and simplify.

a. $f(6)$

b. $f(-6)$

c. $f(r^2)$

Explanation:

Step1: Recall the definition of absolute value

The absolute value function \(|x|\) is defined as \(|x|=

$$\begin{cases}x, & x\geq0\\ -x, & x < 0\end{cases}$$

\). We will use this to simplify the function \(f(x)=\frac{x}{|x|}\) for different values of \(x\).

Step2: Evaluate \(f(6)\)

For \(x = 6\), since \(6>0\), we have \(|6| = 6\). Then \(f(6)=\frac{6}{|6|}=\frac{6}{6}=1\).

Step3: Evaluate \(f(-6)\)

For \(x=-6\), since \(- 6<0\), we have \(|-6|=-(-6) = 6\). Then \(f(-6)=\frac{-6}{|-6|}=\frac{-6}{6}=-1\).

Step4: Evaluate \(f(r^{2})\)

For \(x = r^{2}\), we know that for any real number \(r\), \(r^{2}\geq0\) (because the square of a real number is non - negative). So \(|r^{2}|=r^{2}\) (since \(r^{2}\geq0\)). Then \(f(r^{2})=\frac{r^{2}}{|r^{2}|}=\frac{r^{2}}{r^{2}} = 1\) (assuming \(r
eq0\); if \(r = 0\), the original function \(f(x)=\frac{x}{|x|}\) is not defined at \(x = 0\), but \(r^{2}=0\) when \(r = 0\) and in the context of evaluating \(f(r^{2})\), we consider the domain where the function is defined, so for \(r
eq0\), \(f(r^{2}) = 1\)).

Answer:

a. \(f(6)=1\)

b. \(f(-6)=-1\)

c. \(f(r^{2}) = 1\) (for \(r
eq0\))