To address the question about the exponential function [tex]\( f(x) = 2^x \)[/tex], let's carefully evaluate Geraldine's statements and her conclusion.
Statement 1:
Geraldine states that "as [tex]\( x \)[/tex] increases infinitely, the [tex]\( y \)[/tex]-values are continually doubled for each single increase in [tex]\( x \)[/tex]."
- To verify this, we observe that:
[tex]\[
f(x+1) = 2^{x+1} = 2 \cdot 2^x = 2 \cdot f(x)
\][/tex]
This indicates that for each increase of 1 in [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] (the [tex]\( y \)[/tex]-value) does indeed get doubled. Therefore, Statement 1 is correct.
Statement 2:
Geraldine also states that "as [tex]\( x \)[/tex] decreases infinitely, the [tex]\( y \)[/tex]-values are continually halved for each single decrease in [tex]\( x \)[/tex]."
- To verify this, we observe that:
[tex]\[
f(x-1) = 2^{x-1} = \frac{2^x}{2} = \frac{f(x)}{2}
\][/tex]
This indicates that for each decrease of 1 in [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] (the [tex]\( y \)[/tex]-value) gets halved. Therefore, Statement 2 is also correct.
Conclusion:
Geraldine concludes that "there are no limits within the set of real numbers on the range of this exponential function."
- To clarify this, we need to review the characteristics of the range of the function [tex]\( f(x) = 2^x \)[/tex]. The range of [tex]\( f(x) = 2^x \)[/tex] includes all positive real numbers. This is because as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( 2^x \)[/tex] grows without bound, and as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( 2^x \)[/tex] approaches 0 but never reaches 0. Therefore, the range is all positive real numbers.
Combining these findings:
1. Statement 1 is correct.
2. Statement 2 is correct.
3. Geraldine's conclusion is incorrect because the range of [tex]\( f(x) = 2^x \)[/tex] is limited to the set of positive real numbers, not all real numbers.
Therefore, the best explanation is:
- The conclusion is incorrect because the range is limited to the set of positive real numbers.
