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Sagot :
Let's address the statements and conclusion presented by Geraldine regarding the exponential function [tex]\( f(x) = 2^x \)[/tex].
### Evaluating Statement 1:
1. Statement 1: As [tex]\( x \)[/tex] increases infinitely, the [tex]\( y \)[/tex]-values are continually doubled for each single increase in [tex]\( x \)[/tex].
- An exponential function of the form [tex]\( f(x) = a^x \)[/tex], where [tex]\( a \)[/tex] is a constant greater than 1, satisfies this property. Specifically, for [tex]\( f(x) = 2^x \)[/tex], if [tex]\( x \)[/tex] increases by 1, the value of [tex]\( f(x) \)[/tex] doubles.
- For example, if [tex]\( 2^x \)[/tex] at [tex]\( x=m \)[/tex] results in [tex]\( 2^m \)[/tex], then at [tex]\( x=m+1 \)[/tex], the result is [tex]\( 2^{m+1} = 2 \cdot 2^m \)[/tex], which clearly shows that the value is doubled.
Therefore, Statement 1 is correct.
### Evaluating Statement 2:
2. Statement 2: As [tex]\( x \)[/tex] decreases infinitely, the [tex]\( y \)[/tex]-values are continually halved for each single decrease in [tex]\( x \)[/tex].
- Following similar reasoning, if [tex]\( x \)[/tex] decreases by 1, the value of [tex]\( f(x) \)[/tex] is halved.
- For instance, if [tex]\( 2^x \)[/tex] at [tex]\( x=n \)[/tex] results in [tex]\( 2^n \)[/tex], then at [tex]\( x=n-1 \)[/tex], the result is [tex]\( 2^{n-1} = \frac{1}{2} \cdot 2^n \)[/tex], which clearly shows that the value is halved.
Therefore, Statement 2 is correct.
### Evaluating the Conclusion:
- Geraldine's conclusion is that there are no limits within the set of real numbers on the range of this exponential function.
To analyze this:
- The range of [tex]\( f(x) = 2^x \)[/tex] is all positive real numbers [tex]\( (0, \infty) \)[/tex].
- This means that [tex]\( f(x) \)[/tex] can take any positive real value, no matter how large. However, it can never take on non-positive values (zero or negative numbers).
Therefore, the conclusion that there are 'no limits' is incorrect because there is a limit: [tex]\( f(x) \)[/tex] is confined to positive real numbers.
Thus, the best explanation is:
> The conclusion is incorrect because the range is limited to the set of positive real numbers.
### Evaluating Statement 1:
1. Statement 1: As [tex]\( x \)[/tex] increases infinitely, the [tex]\( y \)[/tex]-values are continually doubled for each single increase in [tex]\( x \)[/tex].
- An exponential function of the form [tex]\( f(x) = a^x \)[/tex], where [tex]\( a \)[/tex] is a constant greater than 1, satisfies this property. Specifically, for [tex]\( f(x) = 2^x \)[/tex], if [tex]\( x \)[/tex] increases by 1, the value of [tex]\( f(x) \)[/tex] doubles.
- For example, if [tex]\( 2^x \)[/tex] at [tex]\( x=m \)[/tex] results in [tex]\( 2^m \)[/tex], then at [tex]\( x=m+1 \)[/tex], the result is [tex]\( 2^{m+1} = 2 \cdot 2^m \)[/tex], which clearly shows that the value is doubled.
Therefore, Statement 1 is correct.
### Evaluating Statement 2:
2. Statement 2: As [tex]\( x \)[/tex] decreases infinitely, the [tex]\( y \)[/tex]-values are continually halved for each single decrease in [tex]\( x \)[/tex].
- Following similar reasoning, if [tex]\( x \)[/tex] decreases by 1, the value of [tex]\( f(x) \)[/tex] is halved.
- For instance, if [tex]\( 2^x \)[/tex] at [tex]\( x=n \)[/tex] results in [tex]\( 2^n \)[/tex], then at [tex]\( x=n-1 \)[/tex], the result is [tex]\( 2^{n-1} = \frac{1}{2} \cdot 2^n \)[/tex], which clearly shows that the value is halved.
Therefore, Statement 2 is correct.
### Evaluating the Conclusion:
- Geraldine's conclusion is that there are no limits within the set of real numbers on the range of this exponential function.
To analyze this:
- The range of [tex]\( f(x) = 2^x \)[/tex] is all positive real numbers [tex]\( (0, \infty) \)[/tex].
- This means that [tex]\( f(x) \)[/tex] can take any positive real value, no matter how large. However, it can never take on non-positive values (zero or negative numbers).
Therefore, the conclusion that there are 'no limits' is incorrect because there is a limit: [tex]\( f(x) \)[/tex] is confined to positive real numbers.
Thus, the best explanation is:
> The conclusion is incorrect because the range is limited to the set of positive real numbers.
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