QUESTION IMAGE
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 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
- the range of ( f(x) ) is all nonzero real numbers, the range of ( g(x) ) is all real numbers except 6
Step1: Recall range of \( f(x)=\frac{1}{x} \)
The function \( f(x)=\frac{1}{x} \) is a reciprocal function. For any real number \( x
eq0 \), \( y = \frac{1}{x}\) will never be \( 0 \) (since if \( \frac{1}{x}=0 \), there's no real \( x \) to satisfy it). So the range of \( f(x) \) is all nonzero real numbers.
Step2: Analyze range of \( g(x)=\frac{6}{x} \)
For \( g(x)=\frac{6}{x} \), similar to \( f(x) \), \( x
eq0 \). If we solve for \( y=\frac{6}{x} \), rearranging gives \( x = \frac{6}{y} \), so \( y
eq0 \) (otherwise, \( x \) would be undefined). Thus, the range of \( g(x) \) is also all nonzero real numbers.
Step3: Evaluate options
- Option 1: Incorrect, as range isn't all real numbers (0 is excluded).
- Option 2: Correct, both have range of all nonzero real numbers.
- Option 3: Incorrect, range of \( g(x) \) isn't excluding 6.
- Option 4: Incorrect, range of \( g(x) \) isn't excluding 6.
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The range of both \( f(x) \) and \( g(x) \) is all nonzero real numbers (the second option).