Given the function [tex][tex]$h(x)=3 \sqrt{x+1}-2$[/tex][/tex], which statement is true about [tex]h(x)[/tex]?

A. The function is decreasing on the interval [tex](-3, \infty)[/tex].
B. The function is decreasing on the interval [tex](-\infty,-1)[/tex].
C. The function is increasing on the interval [tex](-\infty,-3)[/tex].
D. The function is increasing on the interval [tex](-1, \infty)[/tex].



Answer :

To determine the behavior of the function [tex]\( h(x) = 3 \sqrt{x+1} - 2 \)[/tex], we need to analyze in which intervals the function is increasing or decreasing.

1. Understanding the Function [tex]\( h(x) \)[/tex]:
[tex]\[ h(x) = 3 \sqrt{x+1} - 2 \][/tex]
The function involves a square root, which means it is defined for [tex]\( x \geq -1 \)[/tex].

2. Finding the Derivative [tex]\( h'(x) \)[/tex]:
To understand how the function behaves, we need to find its derivative [tex]\( h'(x) \)[/tex]:
[tex]\[ h(x) = 3 (x+1)^{1/2} - 2 \][/tex]
Differentiating this with respect to [tex]\( x \)[/tex]:
[tex]\[ h'(x) = 3 \cdot \frac{1}{2} (x+1)^{-1/2} \cdot 1 = \frac{3}{2} (x+1)^{-1/2} = \frac{3}{2\sqrt{x+1}} \][/tex]

3. Analyzing the Derivative [tex]\( h'(x) \)[/tex]:
The sign of the derivative will tell us whether the function is increasing or decreasing.
[tex]\[ h'(x) = \frac{3}{2\sqrt{x+1}} \][/tex]
For [tex]\( x \geq -1 \)[/tex]:
- The term [tex]\( \sqrt{x+1} \)[/tex] is always positive since [tex]\( x+1 \geq 0 \)[/tex].
- The fraction [tex]\( \frac{3}{2\sqrt{x+1}} \)[/tex] is always positive for [tex]\( x \geq -1 \)[/tex].

Hence, the derivative [tex]\( h'(x) \)[/tex] is positive for all [tex]\( x \geq -1 \)[/tex], indicating that the function is increasing in this interval.

4. Conclusion:
Since the derivative [tex]\( h'(x) \)[/tex] is positive for all [tex]\( x \geq -1 \)[/tex], the function [tex]\( h(x) \)[/tex] is increasing on the interval [tex]\( (-1, \infty) \)[/tex].

Therefore, the correct statement is:

- The function is increasing on the interval [tex]\( (-1, \infty) \)[/tex].

So, the answer is:
[tex]\[ \boxed{\text{The function is increasing on the interval } (-1, \infty).} \][/tex]