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Sagot :
To determine the end behavior of the exponential function [tex]\( f(x) = 2^{x-3} \)[/tex], let's analyze and understand how exponential functions generally behave and specifically how this function behaves.
1. General Behavior of Exponential Functions:
Exponential functions of the form [tex]\( g(x) = a^x \)[/tex], where [tex]\( a > 1 \)[/tex], usually grow rapidly as [tex]\( x \)[/tex] increases. Conversely, as [tex]\( x \)[/tex] decreases, [tex]\( g(x) \)[/tex] approaches 0. This behavior is characteristic of all exponential functions with base [tex]\( a > 1 \)[/tex].
2. Understanding [tex]\( f(x) = 2^{x-3} \)[/tex]:
The function [tex]\( f(x) = 2^{x-3} \)[/tex] can be re-written for clarity in understanding its components:
[tex]\[ f(x) = 2^{x-3} = \left( 2^x \right) \cdot \left( 2^{-3} \right) \][/tex]
Notice that [tex]\( 2^{-3} \)[/tex] is a constant ([tex]\( 2^{-3} = \frac{1}{8} \)[/tex]). This means the function can be expressed as:
[tex]\[ f(x) = \frac{1}{8} \left( 2^x \right) \][/tex]
3. Behavior for Very High [tex]\( x \)[/tex]-Values:
- As [tex]\( x \)[/tex] becomes very large (high [tex]\( x \)[/tex]-values), the term [tex]\( 2^x \)[/tex] grows very quickly towards positive infinity since the base of the exponential, 2, is greater than 1.
- Therefore, the product [tex]\( \frac{1}{8} \cdot 2^x \)[/tex] also moves toward positive infinity because a large positive number multiplied by a positive constant remains a large positive number.
4. Conclusion:
- Given these observations, as [tex]\( x \)[/tex] becomes very large, the function [tex]\( f(x) = 2^{x-3} \)[/tex] moves towards positive infinity.
- This matches the description provided in option A.
Therefore, the correct statement that describes the end behavior of the exponential function [tex]\( f(x) = 2^{x-3} \)[/tex] is:
A. For very high [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward positive infinity.
Thus, the selected answer is:
[tex]\[ \boxed{1} \quad \text{(A)} \][/tex]
1. General Behavior of Exponential Functions:
Exponential functions of the form [tex]\( g(x) = a^x \)[/tex], where [tex]\( a > 1 \)[/tex], usually grow rapidly as [tex]\( x \)[/tex] increases. Conversely, as [tex]\( x \)[/tex] decreases, [tex]\( g(x) \)[/tex] approaches 0. This behavior is characteristic of all exponential functions with base [tex]\( a > 1 \)[/tex].
2. Understanding [tex]\( f(x) = 2^{x-3} \)[/tex]:
The function [tex]\( f(x) = 2^{x-3} \)[/tex] can be re-written for clarity in understanding its components:
[tex]\[ f(x) = 2^{x-3} = \left( 2^x \right) \cdot \left( 2^{-3} \right) \][/tex]
Notice that [tex]\( 2^{-3} \)[/tex] is a constant ([tex]\( 2^{-3} = \frac{1}{8} \)[/tex]). This means the function can be expressed as:
[tex]\[ f(x) = \frac{1}{8} \left( 2^x \right) \][/tex]
3. Behavior for Very High [tex]\( x \)[/tex]-Values:
- As [tex]\( x \)[/tex] becomes very large (high [tex]\( x \)[/tex]-values), the term [tex]\( 2^x \)[/tex] grows very quickly towards positive infinity since the base of the exponential, 2, is greater than 1.
- Therefore, the product [tex]\( \frac{1}{8} \cdot 2^x \)[/tex] also moves toward positive infinity because a large positive number multiplied by a positive constant remains a large positive number.
4. Conclusion:
- Given these observations, as [tex]\( x \)[/tex] becomes very large, the function [tex]\( f(x) = 2^{x-3} \)[/tex] moves towards positive infinity.
- This matches the description provided in option A.
Therefore, the correct statement that describes the end behavior of the exponential function [tex]\( f(x) = 2^{x-3} \)[/tex] is:
A. For very high [tex]\( x \)[/tex]-values, [tex]\( f(x) \)[/tex] moves toward positive infinity.
Thus, the selected answer is:
[tex]\[ \boxed{1} \quad \text{(A)} \][/tex]
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