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. $f(r^2)$

a. $f(6) = \boldsymbol{1}$ (simplify your answer.)

b. $f(-6) = \boldsymbol{4\sqrt{2}}$ (simplify your answer.)

Explanation:

Step1: Evaluate \( f(6) \)

The function is \( f(x)=\frac{x}{|x|} \). For \( x = 6 \), \( |6|=6 \), so \( f(6)=\frac{6}{|6|}=\frac{6}{6}=1 \).

Step2: Evaluate \( f(-6) \)

For \( x=-6 \), \( |-6| = 6 \), so \( f(-6)=\frac{-6}{|-6|}=\frac{-6}{6}=-1 \). (Note: The original answer for b seems incorrect. Let's correct it.)

Step3: Evaluate \( f(r^{2}) \)

Since \( r^{2}\geq0 \) for all real \( r \), when \( r
eq0 \), \( |r^{2}|=r^{2} \), so \( f(r^{2})=\frac{r^{2}}{|r^{2}|}=\frac{r^{2}}{r^{2}} = 1 \). When \( r = 0 \), \( f(0) \) is undefined (but for \( r^{2} \), if \( r = 0 \), \( r^{2}=0 \), and \( f(0) \) is undefined, but in general for \( r^{2}\), we consider \( r\) real, so \( r^{2}\geq0 \), and \( f(r^{2}) = 1 \) for \( r
eq0 \), and undefined at \( r = 0 \), but as a function evaluation for \( r^{2} \), we can say \( f(r^{2})=1 \) (assuming \( r\) is such that \( r^{2}
eq0 \), or in the context of the function, since \( r^{2} \) is non - negative, \( \frac{r^{2}}{|r^{2}|}=1 \) when \( r
eq0 \), and the function is undefined at \( r = 0 \) for \( f(0) \), but for \( f(r^{2}) \), \( r^{2}\) is in the domain as long as \( r^{2}
eq0 \), or we can say for \( r^{2}>0 \), \( f(r^{2}) = 1 \))

Answer:

a. \( f(6)=\boxed{1} \)
b. \( f(-6)=\boxed{-1} \)
c. \( f(r^{2})=\boxed{1} \) (for \( r
eq0 \), and undefined at \( r = 0 \), but in the context of the function evaluation for \( r^{2} \), we have this result)