Question

geraldine is asked to explain the limits on the range of an exponential equation using the function ( f(x) = 2^{x} ). she makes these two statements:

1. as ( x ) increases infinitely, the ( y )-values are continually doubled for each single increase in ( x ).
2. as ( x ) decreases infinitely, the ( y )-values are continually halved for each single decrease in ( x ).

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 geraldines statements and her conclusion?
statement 1 is incorrect because the ( y )-values are increased by 2, not doubled.
statement 2 is incorrect because the ( y )-values are doubled, not halved.
the conclusion is incorrect because the range is limited to the set of integers.
the conclusion is incorrect because the range is limited to the set of positive real numbers.

Explanation:

Step1: Analyze Statement 1

For the function \(y = 2^{x}\), when \(x\) increases by \(1\) (from \(x=n\) to \(x = n + 1\)), \(y_{n}=2^{n}\) and \(y_{n + 1}=2^{n+1}=2\times2^{n}\). So, as \(x\) increases infinitely, the \(y\) - values are continually doubled for each single increase in \(x\). Statement 1 is correct.

Step2: Analyze Statement 2

When \(x\) decreases by \(1\) (from \(x=n\) to \(x=n - 1\)), \(y_{n}=2^{n}\) and \(y_{n-1}=2^{n - 1}=\frac{2^{n}}{2}\). So, as \(x\) decreases infinitely, the \(y\) - values are continually halved for each single decrease in \(x\). Statement 2 is correct.

Step3: Analyze the range of \(y = 2^{x}\)

The exponential function \(y=a^{x}\), where \(a>0,a
eq1\) (here \(a = 2\)) has the range \(y>0\). The set of real - valued \(y\) for \(y = 2^{x}\) is \(y\in(0,\infty)\), which is the set of positive real numbers, not all real numbers.

Answer:

The conclusion is incorrect because the range is limited to the set of positive real numbers.