To determine the behavior of the function [tex]\( h(x) = -2 \sqrt{x-3} \)[/tex], we first need to identify the domain and then analyze the first derivative of the function.
### Step 1: Determine the domain of [tex]\( h(x) \)[/tex]
The function [tex]\( h(x) = -2 \sqrt{x-3} \)[/tex] involves the square root [tex]\( \sqrt{x-3} \)[/tex]. For the square root to be real and defined, the expression inside the square root must be non-negative:
[tex]\[ x - 3 \geq 0 \implies x \geq 3. \][/tex]
So, the domain of [tex]\( h(x) \)[/tex] is [tex]\( [3, \infty) \)[/tex].
### Step 2: Compute the first derivative of [tex]\( h(x) \)[/tex]
We will use differentiation to find the first derivative of [tex]\( h(x) \)[/tex]:
[tex]\[ h(x) = -2 \sqrt{x-3} \][/tex]
Using the chain rule, we have:
[tex]\[ \frac{d}{dx} [\sqrt{x-3}] = \frac{1}{2\sqrt{x-3}} \][/tex]
So,
[tex]\[ h'(x) = \frac{d}{dx} [-2 \sqrt{x-3}] = -2 \cdot \frac{1}{2\sqrt{x-3}} = -\frac{1}{\sqrt{x-3}} \][/tex]
### Step 3: Analyze the first derivative [tex]\( h'(x) \)[/tex]
The expression for the first derivative is:
[tex]\[ h'(x) = -\frac{1}{\sqrt{x-3}} \][/tex]
For [tex]\(x \geq 3\)[/tex], the term [tex]\(\sqrt{x-3}\)[/tex] is always positive. Therefore, [tex]\( -\frac{1}{\sqrt{x-3}} \)[/tex] is always negative for [tex]\( x > 3 \)[/tex].
Since [tex]\( h'(x) \)[/tex] is negative in the interval [tex]\( (3, \infty) \)[/tex], this indicates that [tex]\( h(x) \)[/tex] is a decreasing function in that interval.
### Conclusion
Based on the above analysis, the correct statement that describes [tex]\( h(x) \)[/tex] is:
- The function [tex]\(h(x)\)[/tex] is decreasing on the interval [tex]\( (3, \infty) \)[/tex].
Thus, the correct answer is:
The function [tex]\( h(x) \)[/tex] is decreasing on the interval [tex]\( (3, \infty) \)[/tex].
