determine the intervals over which the function is increasing, decreasing, or constant
f(x)=√x^2-1
The function is increasing over the interval (-∞, -1) and (1, ∞). It is decreasing over the interval (-1, 1). It is constant at x = -1 and x = 1.
To determine the intervals over which the function is increasing, decreasing, or constant, we need to find the derivative of the function and analyze its sign.
Let's find the derivative of the function f(x) = √(x^2 - 1).
First, we denote f(x) as y: y = √(x^2 - 1).
We can rewrite the function as follows: y^2 = (x^2 - 1).
Next, let's differentiate both sides of the equation with respect to x using the chain rule.
2y * dy/dx = 2x.
Simplifying the equation, we have: dy/dx = x / y.
To determine the intervals of increasing, decreasing, or constant behavior, we need to find the regions where the derivative (dy/dx) is positive, negative, or zero.
Since dy/dx = x / y, we need to consider two cases:
1. When y ≠ 0:
- If x > 0, then dy/dx > 0.
- If x < 0, then dy/dx < 0.
2. When y = 0:
In this case, x^2 - 1 = 0.
Solving for x, we find two solutions: x = 1 and x = -1.
Thus, the function is not defined at x = 1 and x = -1.
Based on the above analysis, we can summarize the intervals over which the function f(x) = √(x^2 - 1) is increasing, decreasing, or constant:
1. Increasing Interval: When x > 1, and y ≠ 0.
2. Decreasing Interval: When x < -1, and y ≠ 0.
3. The function is not defined at x = 1 and x = -1, so we exclude these points from the intervals.
Therefore, the function is increasing for x > 1 (excluding x = 1), decreasing for x < -1 (excluding x = -1), and it is not defined at x = 1 and x = -1.
To determine the intervals over which the function is increasing, decreasing, or constant, we need to find the derivative of the function and analyze its sign.
Let's find the derivative of the function first:
f(x) = √(x^2 - 1)
Using the chain rule, the derivative of f(x) is:
f'(x) = d/dx √(x^2 - 1)
= (1/2) * (x^2 - 1)^(-1/2) * (2x)
Simplifying further:
f'(x) = x / √(x^2 - 1)
To determine the intervals, we have to solve the inequality f'(x) > 0 for increasing intervals, f'(x) < 0 for decreasing intervals, and f'(x) = 0 for constant intervals.
Setting f'(x) > 0:
x / √(x^2 - 1) > 0
To have a positive value, either both numerator and denominator should be positive or both should be negative.
Case 1: x > 0 and √(x^2 - 1) > 0
This is true when x > 1.
Case 2: x < 0 and √(x^2 - 1) < 0
This is not possible since the square root will be positive for x < 0.
Therefore, the function is increasing for x > 1.
Setting f'(x) < 0:
x / √(x^2 - 1) < 0
Here, the numerator and denominator should have different signs.
Case 1: x > 0 and √(x^2 - 1) < 0
This is not possible since the square root of a positive number will always be positive.
Case 2: x < 0 and √(x^2 - 1) > 0
This is true when -1 < x < 0.
Therefore, the function is decreasing for -1 < x < 0.
To determine the intervals where the function is constant, we need to find where f'(x) = 0.
Setting f'(x) = 0:
x / √(x^2 - 1) = 0
This equation is only satisfied when x = 0.
Thus, the function is constant at x = 0.
In summary:
- The function is increasing for x > 1.
- The function is decreasing for -1 < x < 0.
- The function is constant at x = 0.