To determine the intervals over which the function [tex]\( h(x) \)[/tex] is decreasing, we need to analyze its behavior and its derivative for both pieces of the function: [tex]\( 2^x \)[/tex] when [tex]\( x < 1 \)[/tex], and [tex]\( \sqrt{x+3} \)[/tex] when [tex]\( x \geq 1 \)[/tex].
First, let's consider the function [tex]\( 2^x \)[/tex] in the interval [tex]\( x < 1 \)[/tex].
1. [tex]\( 2^x \)[/tex] for [tex]\( x < 1 \)[/tex]:
- To understand if this part is increasing or decreasing, we calculate its derivative.
- The derivative of [tex]\( 2^x \)[/tex] is [tex]\( 2^x \ln(2) \)[/tex].
- Since [tex]\( 2^x \)[/tex] is always positive and [tex]\( \ln(2) \)[/tex] is a positive constant, [tex]\( 2^x \ln(2) \)[/tex] is always positive.
- Therefore, [tex]\( 2^x \)[/tex] is always increasing for [tex]\( x < 1 \)[/tex].
Next, let's consider the function [tex]\( \sqrt{x+3} \)[/tex] in the interval [tex]\( x \geq 1 \)[/tex].
2. [tex]\( \sqrt{x+3} \)[/tex] for [tex]\( x \geq 1 \)[/tex]:
- To check if this part is increasing or decreasing, we calculate its derivative.
- The derivative of [tex]\( \sqrt{x+3} \)[/tex] is [tex]\( \frac{1}{2\sqrt{x+3}} \)[/tex].
- Since [tex]\( x+3 \)[/tex] is positive for [tex]\( x \geq 1 \)[/tex], [tex]\( \sqrt{x+3} \)[/tex] is positive and [tex]\( \frac{1}{2\sqrt{x+3}} \)[/tex] is also always positive.
- Therefore, [tex]\( \sqrt{x+3} \)[/tex] is always increasing for [tex]\( x \geq 1 \)[/tex].
From these analyses, we find that both parts of the function are increasing in their respective intervals. Hence, the function [tex]\( h(x) \)[/tex] is never decreasing over its entire domain.
Therefore, the correct answer is:
A. The function is increasing only.
