To determine whether the function [tex]\( h(x) = 3 \sqrt{x+1} - 2 \)[/tex] is increasing or decreasing on certain intervals, we need to analyze the first derivative of [tex]\( h(x) \)[/tex].
### Step 1: Find the first derivative of [tex]\( h(x) \)[/tex]
Starting with the given function:
[tex]\[ h(x) = 3 \sqrt{x+1} - 2 \][/tex]
Rewrite the square root as an exponent to make differentiation easier:
[tex]\[ h(x) = 3 (x + 1)^{1/2} - 2 \][/tex]
Now, differentiate [tex]\( h(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
h'(x) = \frac{d}{dx} \left[ 3 (x+1)^{1/2} - 2 \right]
\][/tex]
[tex]\[
h'(x) = 3 \cdot \frac{1}{2} (x + 1)^{-1/2} \cdot \frac{d}{dx} (x + 1)
\][/tex]
[tex]\[
h'(x) = 3 \cdot \frac{1}{2} (x + 1)^{-1/2}
\][/tex]
[tex]\[
h'(x) = \frac{3}{2} (x + 1)^{-1/2}
\][/tex]
[tex]\[
h'(x) = \frac{3}{2 \sqrt{x + 1}}
\][/tex]
### Step 2: Analyze the sign of the first derivative
The sign of [tex]\( h'(x) \)[/tex] indicates whether [tex]\( h(x) \)[/tex] is increasing or decreasing. Note that:
[tex]\[ \frac{3}{2 \sqrt{x + 1}} \][/tex]
The expression [tex]\( \sqrt{x + 1} \)[/tex] is always positive for [tex]\( x > -1 \)[/tex], which makes [tex]\( h'(x) \)[/tex] positive for [tex]\( x > -1 \)[/tex]. Therefore, [tex]\( h(x) \)[/tex] is increasing on the interval [tex]\( (-1, \infty) \)[/tex].
### Step 3: Determine the valid domain of the function
The function [tex]\( h(x) \)[/tex] involves a square root, which is defined only for [tex]\( x + 1 \geq 0 \)[/tex], or equivalently:
[tex]\[ x \geq -1 \][/tex]
Thus, the domain of [tex]\( h(x) \)[/tex] is [tex]\( [-1, \infty) \)[/tex].
### Conclusion
Given that [tex]\( h'(x) \)[/tex] is positive [tex]\( (i.e., \, \frac{3}{2 \sqrt{x + 1}} > 0 ) \)[/tex] for [tex]\( x > -1 \)[/tex] and the valid domain of [tex]\( h(x) \)[/tex] is [tex]\( [-1, \infty) \)[/tex], we can conclude:
- The function [tex]\( h(x) \)[/tex] is increasing on the interval [tex]\( (-1, \infty) \)[/tex].
Among the provided answer choices, the correct statement is:
- The function is increasing on the interval [tex]\((-1, \infty)\)[/tex].
