Answer :
To determine the interval in which the radical function [tex]\( f(x) = \sqrt{x^2 + 2x - 15} \)[/tex] is increasing, we follow these steps:
### 1. Find the Derivative of [tex]\( f(x) \)[/tex]
The first step involves finding the derivative of the given function [tex]\( f(x) \)[/tex].
Given:
[tex]\[ f(x) = \sqrt{x^2 + 2x - 15} \][/tex]
We need to find [tex]\( f'(x) \)[/tex] (the derivative of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex]).
### 2. Determine Critical Points
Critical points occur where [tex]\( f'(x) = 0 \)[/tex] or [tex]\( f'(x) \)[/tex] is undefined. Solving for these points will help us understand where the function changes its behavior.
From the given result:
[tex]\[ f'(x) = \frac{x + 1}{\sqrt{x^2 + 2x - 15}} \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ \frac{x + 1}{\sqrt{x^2 + 2x - 15}} = 0 \][/tex]
This implies:
[tex]\[ x + 1 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
So, [tex]\( x = -1 \)[/tex] is a critical point.
### 3. Analyze Intervals Around Critical Points
To understand where the function [tex]\( f(x) \)[/tex] is increasing, we need to analyze the sign of [tex]\( f'(x) \)[/tex] in the intervals around the critical point [tex]\( x = -1 \)[/tex].
The second part of the given result indicates an interval of interest:
[tex]\[ \text{Interval} = (-\infty, 3) \][/tex]
### 4. Determine Behavior of [tex]\( f(x) \)[/tex] in Given Intervals
Since the function is defined and we are interested in intervals where it is increasing, we analyze:
- For [tex]\( x < -1 \)[/tex], select a test point [tex]\( x = -2 \)[/tex]:
[tex]\[ f'(-2) = \frac{-2 + 1}{\sqrt{(-2)^2 + 2(-2) - 15}} = \frac{-1}{\sqrt{4 - 4 - 15}} \][/tex]
The denominator cannot be evaluated as the expression inside the square root is negative, so the function is not real in this part of the interval.
- For [tex]\( -1 < x < 3 \)[/tex], select a test point [tex]\( x = 0 \)[/tex]:
[tex]\[ f'(0) = \frac{0 + 1}{\sqrt{0^2 + 2(0) - 15}} = \frac{1}{\sqrt{-15}} \][/tex]
Again, the expression inside the square root is negative for these values, thus, the evaluation is invalid.
- For [tex]\( x > 3 \)[/tex], select a test point [tex]\( x = 4 \)[/tex]:
[tex]\[ f'(4) = \frac{4 + 1}{\sqrt{4^2 + 2(4) - 15}} = \frac{5}{\sqrt{16 + 8 - 15}} = \frac{5}{\sqrt{9}} = \frac{5}{3} \][/tex]
The result is positive, so [tex]\( f(x) \)[/tex] is increasing in this interval.
### Conclusion
Considering [tex]\( f(x) = \sqrt{x^2 + 2x - 15} \)[/tex] and analyzing [tex]\( f'(x) \)[/tex], we conclude that [tex]\( f(x) \)[/tex] is increasing in the interval:
[tex]\[ [3, \infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{[3, \infty)} \][/tex]
### 1. Find the Derivative of [tex]\( f(x) \)[/tex]
The first step involves finding the derivative of the given function [tex]\( f(x) \)[/tex].
Given:
[tex]\[ f(x) = \sqrt{x^2 + 2x - 15} \][/tex]
We need to find [tex]\( f'(x) \)[/tex] (the derivative of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex]).
### 2. Determine Critical Points
Critical points occur where [tex]\( f'(x) = 0 \)[/tex] or [tex]\( f'(x) \)[/tex] is undefined. Solving for these points will help us understand where the function changes its behavior.
From the given result:
[tex]\[ f'(x) = \frac{x + 1}{\sqrt{x^2 + 2x - 15}} \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ \frac{x + 1}{\sqrt{x^2 + 2x - 15}} = 0 \][/tex]
This implies:
[tex]\[ x + 1 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
So, [tex]\( x = -1 \)[/tex] is a critical point.
### 3. Analyze Intervals Around Critical Points
To understand where the function [tex]\( f(x) \)[/tex] is increasing, we need to analyze the sign of [tex]\( f'(x) \)[/tex] in the intervals around the critical point [tex]\( x = -1 \)[/tex].
The second part of the given result indicates an interval of interest:
[tex]\[ \text{Interval} = (-\infty, 3) \][/tex]
### 4. Determine Behavior of [tex]\( f(x) \)[/tex] in Given Intervals
Since the function is defined and we are interested in intervals where it is increasing, we analyze:
- For [tex]\( x < -1 \)[/tex], select a test point [tex]\( x = -2 \)[/tex]:
[tex]\[ f'(-2) = \frac{-2 + 1}{\sqrt{(-2)^2 + 2(-2) - 15}} = \frac{-1}{\sqrt{4 - 4 - 15}} \][/tex]
The denominator cannot be evaluated as the expression inside the square root is negative, so the function is not real in this part of the interval.
- For [tex]\( -1 < x < 3 \)[/tex], select a test point [tex]\( x = 0 \)[/tex]:
[tex]\[ f'(0) = \frac{0 + 1}{\sqrt{0^2 + 2(0) - 15}} = \frac{1}{\sqrt{-15}} \][/tex]
Again, the expression inside the square root is negative for these values, thus, the evaluation is invalid.
- For [tex]\( x > 3 \)[/tex], select a test point [tex]\( x = 4 \)[/tex]:
[tex]\[ f'(4) = \frac{4 + 1}{\sqrt{4^2 + 2(4) - 15}} = \frac{5}{\sqrt{16 + 8 - 15}} = \frac{5}{\sqrt{9}} = \frac{5}{3} \][/tex]
The result is positive, so [tex]\( f(x) \)[/tex] is increasing in this interval.
### Conclusion
Considering [tex]\( f(x) = \sqrt{x^2 + 2x - 15} \)[/tex] and analyzing [tex]\( f'(x) \)[/tex], we conclude that [tex]\( f(x) \)[/tex] is increasing in the interval:
[tex]\[ [3, \infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{[3, \infty)} \][/tex]
