To determine the behavior of the function [tex]\( h(x) = -2 \sqrt{x+3} - 1 \)[/tex], we need to analyze its derivative. Here’s how we can approach it step-by-step:
1. Determine the domain of [tex]\( h(x) \)[/tex]:
[tex]\[
x + 3 \geq 0 \implies x \geq -3
\][/tex]
Hence, the domain of [tex]\( h(x) \)[/tex] is [tex]\( x \in [-3, \infty) \)[/tex].
2. Find the derivative of [tex]\( h(x) \)[/tex]:
[tex]\[
h(x) = -2 \sqrt{x+3} - 1
\][/tex]
Using the chain rule:
[tex]\[
h'(x) = -2 \cdot \frac{1}{2} \cdot (x+3)^{-1/2} \cdot 1 = -\frac{1}{\sqrt{x+3}}
\][/tex]
3. Analyze the sign of the derivative [tex]\( h'(x) \)[/tex]:
[tex]\[
h'(x) = -\frac{1}{\sqrt{x+3}}
\][/tex]
Since the square root function [tex]\( \sqrt{x+3} \)[/tex] is always positive for [tex]\( x \geq -3 \)[/tex], [tex]\( -\frac{1}{\sqrt{x+3}} \)[/tex] is always negative in the domain of [tex]\( h(x) \)[/tex].
4. Conclusion about the monotonicity of [tex]\( h(x) \)[/tex]:
Since [tex]\( h'(x) \)[/tex] is always negative for [tex]\( x \geq -3 \)[/tex], the function [tex]\( h(x) \)[/tex] is always decreasing on its entire domain [tex]\([-3, \infty)\)[/tex].
Therefore, the correct statement about the function [tex]\( h(x) \)[/tex] is:
The function is decreasing on the interval [tex]\(( -3, \infty)\)[/tex].
