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Sagot :
To determine if the exponential function [tex]\( f(x) = a^x \)[/tex], where [tex]\( a > 1 \)[/tex], is increasing or decreasing over its entire domain, let's go through a step-by-step analysis.
1. Understanding the function: An exponential function of the form [tex]\( f(x) = a^x \)[/tex] involves a constant base [tex]\( a \)[/tex] raised to a variable exponent [tex]\( x \)[/tex]. Here, we are given that [tex]\( a > 1 \)[/tex].
2. Behavior for different values of [tex]\( x \)[/tex]:
- If [tex]\( x \)[/tex] is a positive number, say [tex]\( x = 2 \)[/tex], then [tex]\( a^x = a^2 \)[/tex]. Since [tex]\( a > 1 \)[/tex], [tex]\( a^2 \)[/tex] will be larger than [tex]\( a \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is larger as [tex]\( x \)[/tex] increases when [tex]\( x \)[/tex] is positive.
- If [tex]\( x \)[/tex] is zero, [tex]\( a^0 = 1 \)[/tex] because any number raised to the power of zero is 1.
- If [tex]\( x \)[/tex] is a negative number, say [tex]\( x = -1 \)[/tex], then [tex]\( a^x = a^{-1} = \frac{1}{a} \)[/tex]. Since [tex]\( a > 1 \)[/tex], [tex]\(\frac{1}{a} \)[/tex] will be less than 1. As [tex]\( x \)[/tex] becomes more negative, [tex]\( a^x \)[/tex] will continue to decrease but will always be positive and approaching zero.
3. Overall trend:
- When [tex]\( x \)[/tex] increases (i.e., becomes more positive), [tex]\( f(x) \)[/tex] increases because the base [tex]\( a \)[/tex] is being raised to larger and larger powers.
- When [tex]\( x \)[/tex] decreases (i.e., becomes more negative), [tex]\( f(x) \)[/tex] also decreases but remains positive.
4. Observation from the derivative:
- The derivative of [tex]\( f(x) = a^x \)[/tex], where [tex]\( a > 1 \)[/tex], is [tex]\( f'(x) = a^x \ln(a) \)[/tex].
- Since [tex]\( a > 1 \)[/tex], [tex]\( \ln(a) \)[/tex] (the natural logarithm of [tex]\( a \)[/tex]) is positive.
- Therefore, [tex]\( f'(x) \)[/tex] is positive for all [tex]\( x \)[/tex].
Since the derivative [tex]\( f'(x) \)[/tex] is positive for the entire domain of [tex]\( x \)[/tex], it indicates that [tex]\( f(x) = a^x \)[/tex] is an increasing function for all values of [tex]\( x \)[/tex].
Therefore, the correct answer is:
B. increasing
1. Understanding the function: An exponential function of the form [tex]\( f(x) = a^x \)[/tex] involves a constant base [tex]\( a \)[/tex] raised to a variable exponent [tex]\( x \)[/tex]. Here, we are given that [tex]\( a > 1 \)[/tex].
2. Behavior for different values of [tex]\( x \)[/tex]:
- If [tex]\( x \)[/tex] is a positive number, say [tex]\( x = 2 \)[/tex], then [tex]\( a^x = a^2 \)[/tex]. Since [tex]\( a > 1 \)[/tex], [tex]\( a^2 \)[/tex] will be larger than [tex]\( a \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] is larger as [tex]\( x \)[/tex] increases when [tex]\( x \)[/tex] is positive.
- If [tex]\( x \)[/tex] is zero, [tex]\( a^0 = 1 \)[/tex] because any number raised to the power of zero is 1.
- If [tex]\( x \)[/tex] is a negative number, say [tex]\( x = -1 \)[/tex], then [tex]\( a^x = a^{-1} = \frac{1}{a} \)[/tex]. Since [tex]\( a > 1 \)[/tex], [tex]\(\frac{1}{a} \)[/tex] will be less than 1. As [tex]\( x \)[/tex] becomes more negative, [tex]\( a^x \)[/tex] will continue to decrease but will always be positive and approaching zero.
3. Overall trend:
- When [tex]\( x \)[/tex] increases (i.e., becomes more positive), [tex]\( f(x) \)[/tex] increases because the base [tex]\( a \)[/tex] is being raised to larger and larger powers.
- When [tex]\( x \)[/tex] decreases (i.e., becomes more negative), [tex]\( f(x) \)[/tex] also decreases but remains positive.
4. Observation from the derivative:
- The derivative of [tex]\( f(x) = a^x \)[/tex], where [tex]\( a > 1 \)[/tex], is [tex]\( f'(x) = a^x \ln(a) \)[/tex].
- Since [tex]\( a > 1 \)[/tex], [tex]\( \ln(a) \)[/tex] (the natural logarithm of [tex]\( a \)[/tex]) is positive.
- Therefore, [tex]\( f'(x) \)[/tex] is positive for all [tex]\( x \)[/tex].
Since the derivative [tex]\( f'(x) \)[/tex] is positive for the entire domain of [tex]\( x \)[/tex], it indicates that [tex]\( f(x) = a^x \)[/tex] is an increasing function for all values of [tex]\( x \)[/tex].
Therefore, the correct answer is:
B. increasing
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