Let's examine the function [tex]\( h(x) = 3\sqrt{x+2} \)[/tex] to determine its behavior.
1. Find the derivative [tex]\(h'(x)\)[/tex]:
To determine where the function is increasing or decreasing, we need to find the first derivative of [tex]\( h(x) \)[/tex].
[tex]\[
h(x) = 3\sqrt{x+2}
\][/tex]
To find the derivative, we use the chain rule. Recall that the derivative of [tex]\( \sqrt{u} \)[/tex] with respect to [tex]\( u \)[/tex] is [tex]\( \frac{1}{2\sqrt{u}} \)[/tex], and let [tex]\( u = x + 2 \)[/tex].
[tex]\[
h'(x) = 3 \cdot \frac{1}{2\sqrt{x+2}} \cdot (1) = \frac{3}{2\sqrt{x+2}}
\][/tex]
2. Simplified first derivative:
The simplified first derivative is:
[tex]\[
h'(x) = \frac{3}{2\sqrt{x+2}}
\][/tex]
3. Determine where the function is increasing:
The function [tex]\( h(x) \)[/tex] will be increasing where [tex]\( h'(x) > 0 \)[/tex].
[tex]\[
\frac{3}{2\sqrt{x+2}} > 0
\][/tex]
Since the square root function [tex]\(\sqrt{x+2}\)[/tex] is always positive for [tex]\( x > -2 \)[/tex], [tex]\(h'(x)\)[/tex] is also always positive for [tex]\( x > -2 \)[/tex].
4. Domain of [tex]\( h(x) \)[/tex]:
The domain of [tex]\( h(x) \)[/tex] is determined by the requirement that the expression under the square root must be non-negative:
[tex]\[
x + 2 \geq 0 \implies x \geq -2
\][/tex]
5. Conclusion on intervals where [tex]\( h(x) \)[/tex] is increasing:
Because [tex]\( h'(x) \)[/tex] is always positive for [tex]\( x > -2 \)[/tex], the function [tex]\( h(x) \)[/tex] is increasing on its domain [tex]\( (-2, \infty) \)[/tex].
Therefore, the correct statement describing [tex]\( h(x) = 3\sqrt{x+2} \)[/tex] is:
- The function [tex]\( h(x) \)[/tex] is increasing on the interval [tex]\( (-2, \infty) \)[/tex].
