IDNLearn.com makes it easy to find accurate answers to your questions. Join our Q&A platform to access reliable and detailed answers from experts in various fields.
Sagot :
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]
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]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for trusting IDNLearn.com. We’re dedicated to providing accurate answers, so visit us again for more solutions.