To analyze the effects on the function \( g(x) = \sqrt{2x - 1} + 2 \) when it changes to \( f(x) = \sqrt{2x - 1} + 10 \), let's consider the following aspects: the domain and the range.
Domain
The domain of a function is determined by the values of \( x \) for which the expression under the square root is non-negative.
For \( g(x) \):
- We set \( 2x - 1 \geq 0 \), which leads to \( 2x \geq 1 \) or \( x \geq \frac{1}{2} \).
- Thus, the domain of \( g(x) \) is \( [\frac{1}{2}, \infty) \).
For \( f(x) \):
- The domain remains the same: \( 2x - 1 \geq 0 \) leads to the same conclusion \( x \geq \frac{1}{2} \).
- Thus, the domain of \( f(x) \) is also \( [\frac{1}{2}, \infty) \).
Range
The range of the function can be determined by analyzing the outputs of the functions based on their minimum values.
For \( g(x) \):
- The minimum value occurs when \( x = \frac{1}{2} \): \[ g\left(\frac{1}{2}\right) = \sqrt{2(\frac{1}{2}) - 1} + 2 = \sqrt{0} + 2 = 2 \]
- Therefore, the range of \( g(x) \) is \( [2, \infty) \).
For \( f(x) \):
- The minimum value occurs at the same \( x = \frac{1}{2} \): \[ f\left(\frac{1}{2}\right) = \sqrt{2(\frac{1}{2}) - 1} + 10 = \sqrt{0} + 10 = 10 \]
- Therefore, the range of \( f(x) \) is \( [10, \infty) \).
Summary of Changes
- Domain: The domain remains the same \( [\frac{1}{2}, \infty) \).
- Range: The range changes from \( [2, \infty) \) for \( g(x) \) to \( [10, \infty) \) for \( f(x) \).
Correct Response
Based on this analysis, the correct option describing the effect is:
The range changes from \( [2, \infty) \) to \( [10, \infty) \).