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Given the function [tex]$h(x)=3 \sqrt{x+1}-2$[/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].


Sagot :

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].