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Sagot :
To determine which of the given statements is true for the functions [tex]\( f(x) = \log_2(2x) \)[/tex] and [tex]\( g(x) = 2^x - 3 \)[/tex], let's analyze the properties of these functions:
1. Behavior on the Interval [tex]\((-∞, 1)\)[/tex]:
- For [tex]\( f(x) = \log_2(2x) \)[/tex]:
- The derivative is [tex]\( f'(x) = \frac{1}{x\ln(2)} \)[/tex].
- Evaluating the derivative at various points indicates the nature of increase or decrease.
- As [tex]\( x \to -∞ \)[/tex], [tex]\( f'(x) \)[/tex] approaches [tex]\( 0 \)[/tex].
- At [tex]\( x = 1 \)[/tex], [tex]\( f'(1) = \frac{1}{\ln(2)} \)[/tex].
- For [tex]\( g(x) = 2^x - 3 \)[/tex]:
- The derivative is [tex]\( g'(x) = 2^x \ln(2) \)[/tex].
- As [tex]\( x \to -∞ \)[/tex], [tex]\( g'(x) \)[/tex] approaches [tex]\( 0 \)[/tex].
- At [tex]\( x = 1 \)[/tex], [tex]\( g'(1) = 2 \ln(2) \)[/tex].
Since [tex]\( f'(x) > 0 \)[/tex] and [tex]\( g'(x) > 0 \)[/tex] for these values, neither [tex]\( f(x) \)[/tex] nor [tex]\( g(x) \)[/tex] decreases on the interval [tex]\((-∞, 1)\)[/tex].
2. Range:
- For [tex]\( f(x) = \log_2(2x) \)[/tex]:
- The range is [tex]\( (-∞, ∞) \)[/tex].
- For [tex]\( g(x) = 2^x - 3 \)[/tex]:
- The range is [tex]\( (-3, ∞) \)[/tex].
Since the ranges are different, both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] do not have the same range of [tex]\( (-∞, 0] \)[/tex].
3. x-intercepts:
- For [tex]\( f(x) = \log_2(2x) \)[/tex]:
- Set [tex]\( \log_2(2x) = 0 \)[/tex], solving for [tex]\( x \)[/tex] gives [tex]\( x = \frac{1}{2} \)[/tex].
- For [tex]\( g(x) = 2^x - 3 \)[/tex]:
- Set [tex]\( 2^x - 3 = 0 \)[/tex], solving for [tex]\( x \)[/tex] gives [tex]\( x = \log_2(3) \)[/tex].
The [tex]\( x \)[/tex]-intercepts do not coincide, so both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] do not have the same [tex]\( x \)[/tex]-intercept of [tex]\((0,0)\)[/tex].
4. Behavior on the Interval [tex]\((0, ∞)\)[/tex]:
- For [tex]\( f(x) = \log_2(2x) \)[/tex]:
- The derivative is [tex]\( f'(x) = \frac{1}{x \ln(2)} \)[/tex], which is positive for [tex]\( x > 0 \)[/tex].
- As [tex]\( x \to ∞ \)[/tex], [tex]\( f'(x) \)[/tex] approaches [tex]\( 0 \)[/tex] from the positive side.
- For [tex]\( g(x) = 2^x - 3 \)[/tex]:
- The derivative is [tex]\( g'(x) = 2^x \ln(2) \)[/tex], which is positive for all [tex]\( x \)[/tex].
- As [tex]\( x \to ∞ \)[/tex], [tex]\( g'(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
Since [tex]\( f'(x) > 0 \)[/tex] and [tex]\( g'(x) > 0 \)[/tex] for [tex]\( x \in (0, ∞) \)[/tex], both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] increase on the interval [tex]\( (0, ∞) \)[/tex].
The true statement is: Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] increase on the interval [tex]\( (0, \infty) \)[/tex].
1. Behavior on the Interval [tex]\((-∞, 1)\)[/tex]:
- For [tex]\( f(x) = \log_2(2x) \)[/tex]:
- The derivative is [tex]\( f'(x) = \frac{1}{x\ln(2)} \)[/tex].
- Evaluating the derivative at various points indicates the nature of increase or decrease.
- As [tex]\( x \to -∞ \)[/tex], [tex]\( f'(x) \)[/tex] approaches [tex]\( 0 \)[/tex].
- At [tex]\( x = 1 \)[/tex], [tex]\( f'(1) = \frac{1}{\ln(2)} \)[/tex].
- For [tex]\( g(x) = 2^x - 3 \)[/tex]:
- The derivative is [tex]\( g'(x) = 2^x \ln(2) \)[/tex].
- As [tex]\( x \to -∞ \)[/tex], [tex]\( g'(x) \)[/tex] approaches [tex]\( 0 \)[/tex].
- At [tex]\( x = 1 \)[/tex], [tex]\( g'(1) = 2 \ln(2) \)[/tex].
Since [tex]\( f'(x) > 0 \)[/tex] and [tex]\( g'(x) > 0 \)[/tex] for these values, neither [tex]\( f(x) \)[/tex] nor [tex]\( g(x) \)[/tex] decreases on the interval [tex]\((-∞, 1)\)[/tex].
2. Range:
- For [tex]\( f(x) = \log_2(2x) \)[/tex]:
- The range is [tex]\( (-∞, ∞) \)[/tex].
- For [tex]\( g(x) = 2^x - 3 \)[/tex]:
- The range is [tex]\( (-3, ∞) \)[/tex].
Since the ranges are different, both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] do not have the same range of [tex]\( (-∞, 0] \)[/tex].
3. x-intercepts:
- For [tex]\( f(x) = \log_2(2x) \)[/tex]:
- Set [tex]\( \log_2(2x) = 0 \)[/tex], solving for [tex]\( x \)[/tex] gives [tex]\( x = \frac{1}{2} \)[/tex].
- For [tex]\( g(x) = 2^x - 3 \)[/tex]:
- Set [tex]\( 2^x - 3 = 0 \)[/tex], solving for [tex]\( x \)[/tex] gives [tex]\( x = \log_2(3) \)[/tex].
The [tex]\( x \)[/tex]-intercepts do not coincide, so both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] do not have the same [tex]\( x \)[/tex]-intercept of [tex]\((0,0)\)[/tex].
4. Behavior on the Interval [tex]\((0, ∞)\)[/tex]:
- For [tex]\( f(x) = \log_2(2x) \)[/tex]:
- The derivative is [tex]\( f'(x) = \frac{1}{x \ln(2)} \)[/tex], which is positive for [tex]\( x > 0 \)[/tex].
- As [tex]\( x \to ∞ \)[/tex], [tex]\( f'(x) \)[/tex] approaches [tex]\( 0 \)[/tex] from the positive side.
- For [tex]\( g(x) = 2^x - 3 \)[/tex]:
- The derivative is [tex]\( g'(x) = 2^x \ln(2) \)[/tex], which is positive for all [tex]\( x \)[/tex].
- As [tex]\( x \to ∞ \)[/tex], [tex]\( g'(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
Since [tex]\( f'(x) > 0 \)[/tex] and [tex]\( g'(x) > 0 \)[/tex] for [tex]\( x \in (0, ∞) \)[/tex], both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] increase on the interval [tex]\( (0, ∞) \)[/tex].
The true statement is: Both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] increase on the interval [tex]\( (0, \infty) \)[/tex].
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