To determine over which interval the function [tex]\( h(x) \)[/tex] is decreasing, let's analyze each piece of the function separately.
The function [tex]\( h(x) \)[/tex] is defined as follows:
[tex]\[ h(x)=\left\{\begin{array}{ll}
2^x, & x<1 \\
\sqrt{x+3}, & x \geq 1
\end{array}\right. \][/tex]
### Analysis of Each Piece
#### For [tex]\( x < 1 \)[/tex]:
The function given is [tex]\( 2^x \)[/tex].
- The exponential function [tex]\( 2^x \)[/tex] is known to be always increasing for all values of [tex]\( x \)[/tex]. This means as [tex]\( x \)[/tex] increases, [tex]\( 2^x \)[/tex] also increases.
- Since this function is increasing for all [tex]\( x \)[/tex], it is also increasing on the interval [tex]\( (-\infty, 1) \)[/tex].
#### For [tex]\( x \geq 1 \)[/tex]:
The function given is [tex]\( \sqrt{x+3} \)[/tex].
- The square root function [tex]\( \sqrt{x+3} \)[/tex] is defined for [tex]\( x \geq -3 \)[/tex].
- Since [tex]\( x \geq 1 \)[/tex], [tex]\( \sqrt{x+3} \)[/tex] simplifies to a square root function that is always increasing. As [tex]\( x \)[/tex] increases from 1 onward, the value of [tex]\( \sqrt{x+3} \)[/tex] also increases.
- Therefore, this function is increasing on the interval [tex]\( [1, \infty) \)[/tex].
### Conclusion
Since both parts of the function [tex]\( h(x) \)[/tex] are increasing in their respective domains ([tex]\( 2^x \)[/tex] for [tex]\( x < 1 \)[/tex] and [tex]\( \sqrt{x+3} \)[/tex] for [tex]\( x \geq 1 \)[/tex]), there is no interval over which [tex]\( h(x) \)[/tex] is decreasing.
Thus, the correct answer is:
A. The function is increasing only.
