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Analyze the effect on the domain and range of \( f(x)=\ln (x) \) when \( f(x) \) is replaced by \( a f(x) \) where \( a<0 \). Option \#1: Neither the domain nor the range change. Option \#2: The domain changes, but the range does not. Option \#3: The range changes, but the domain does not. Option \#4: Both the domain and the range change. (1 point) Option \# describes the effect on the domain and range of \( f(x)=\ln (x) \) when \( f(x) \) is replaced

Ask by Schmidt Watkins. in the United States
Feb 26,2025

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Neither the domain nor the range of \( f(x) = \ln(x) \) changes when multiplied by a negative constant \( a \).

Solución

The function f(x) = ln(x) has a domain of (0, ∞) and a range of (–∞, ∞). When you multiply by a constant a where a < 0, the operation becomes g(x) = a·ln(x). Multiplying by a constant does not affect the x-values for which ln(x) is defined. Therefore, the domain remains (0, ∞). Since ln(x) can take any real value, multiplying by any nonzero constant (even a negative one) will still produce all real numbers (although the output will be "stretched" and reflected). Thus, the range remains (–∞, ∞). So, the correct option is Option #1: Neither the domain nor the range change.

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Extra Insights

The correct option is Option #3: The range changes, but the domain does not. Here's why: The domain of \( f(x) = \ln(x) \) is \( x > 0 \), which remains unchanged even when \( f(x) \) is replaced by \( a f(x) \) where \( a < 0 \). However, the range of \( f(x) \) is all real numbers, but when \( a < 0 \), the range is inverted to all negative real numbers. Thus, the effect is that while the domain stays the same, the negative multiplier alters the range.

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