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Geraldine is asked to explain the limits on the range of an exponential equation using the function [tex]f(x)=2^x[/tex].

She makes these two statements:
1. As [tex]x[/tex] increases infinitely, the [tex]y[/tex]-values are continually doubled for each single increase in [tex]x[/tex].
2. As [tex]x[/tex] decreases infinitely, the [tex]y[/tex]-values are continually halved for each single decrease in [tex]x[/tex].

She concludes that there are no limits within the set of real numbers on the range of this exponential function.

Which best explains the accuracy of Geraldine's statements and her conclusion?

A. Statement 1 is incorrect because the [tex]y[/tex]-values are increased by 2, not doubled.
B. Statement 2 is incorrect because the [tex]y[/tex]-values are doubled, not halved.
C. The conclusion is incorrect because the range is limited to the set of integers.
D. The conclusion is incorrect because the range is limited to the set of positive real numbers.

Sagot :

GinnyAnswer
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.
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