Question

how does the range of $g(x) = \frac{6}{x}$ compare with the range of the parent function $f(x) = \frac{1}{x}$?
the range of $f(x)$ is all nonzero real numbers, the range of $g(x)$ is all real numbers except 6
the range of both $f(x)$ and $g(x)$ is all real numbers
the range of both $f(x)$ and $g(x)$ is all nonzero real numbers
the range of $f(x)$ is all real numbers, the range of $g(x)$ is all real numbers except 6

Explanation:

Step1: Recall Range of Reciprocal Function

The parent function \( f(x) = \frac{1}{x} \) is a reciprocal function. For \( f(x) = \frac{1}{x} \), as \( x \) approaches \( 0 \) from the positive side, \( f(x) \) approaches \( +\infty \); as \( x \) approaches \( 0 \) from the negative side, \( f(x) \) approaches \( -\infty \). Also, \( f(x) \) can never be \( 0 \) (since \( \frac{1}{x} = 0 \) has no solution). So the range of \( f(x) \) is all nonzero real numbers.

Step2: Analyze Range of \( g(x) = \frac{6}{x} \)

For \( g(x) = \frac{6}{x} \), let's consider the possible values of \( g(x) \). If we solve for \( x \) in terms of \( y \) (where \( y = g(x) \)), we get \( y = \frac{6}{x} \implies x = \frac{6}{y} \). Here, \( x \) must be a real number (and \( x
eq 0 \)), so \( y \) cannot be \( 0 \) (because if \( y = 0 \), then \( x = \frac{6}{0} \) is undefined). Also, as \( x \) approaches \( 0 \) from the positive side, \( y = \frac{6}{x} \) approaches \( +\infty \); as \( x \) approaches \( 0 \) from the negative side, \( y \) approaches \( -\infty \). So the range of \( g(x) \) is also all nonzero real numbers.

Step3: Compare Ranges

The range of \( f(x) = \frac{1}{x} \) is all nonzero real numbers, and the range of \( g(x) = \frac{6}{x} \) is also all nonzero real numbers. So the range of both functions is all nonzero real numbers.

Answer:

The range of both \( f(x) \) and \( g(x) \) is all nonzero real numbers (the third option, assuming the third option is "The range of both \( f(x) \) and \( g(x) \) is all nonzero real numbers").