To determine the interval of the domain over which the function [tex]\( h(x) \)[/tex] is decreasing, let’s analyze the behavior of [tex]\( h(x) \)[/tex] in both defined intervals separately.
The function [tex]\( h(x) \)[/tex] is defined as:
[tex]\[
h(x) = \left\{
\begin{array}{ll}
2^x, & x < 1 \\
\sqrt{x + 3}, & x \geq 1
\end{array}
\right.
\][/tex]
### Interval [tex]\( x < 1 \)[/tex]:
For [tex]\( x < 1 \)[/tex], [tex]\( h(x) = 2^x \)[/tex].
- The function [tex]\( 2^x \)[/tex] is an exponential function with base 2.
- Exponential functions with bases greater than 1 are always increasing.
- Therefore, [tex]\( h(x) \)[/tex] is increasing for [tex]\( x < 1 \)[/tex].
### Interval [tex]\( x \geq 1 \)[/tex]:
For [tex]\( x \geq 1 \)[/tex], [tex]\( h(x) = \sqrt{x + 3} \)[/tex].
- The function [tex]\( \sqrt{x + 3} \)[/tex] is a square root function shifted horizontally by -3.
- Square root functions are increasing for their domains because the derivative of [tex]\( \sqrt{x} \)[/tex] (which is [tex]\( \frac{1}{2\sqrt{x}} \)[/tex] for [tex]\( x \geq 0 \)[/tex]) is positive.
- Therefore, [tex]\( \sqrt{x + 3} \)[/tex] is also increasing for [tex]\( x \geq 1 \)[/tex].
### Conclusion:
Both pieces of the piecewise function [tex]\( h(x) \)[/tex] are increasing in their respective intervals. Consequently, the function [tex]\( h(x) \)[/tex] is increasing throughout its entire domain and does not have any interval where it is decreasing.
Hence, the correct answer is:
D. The function is increasing only.
