Answer:
So, for x > 1, the function f(x) = 4/(x^2 - 2x - 3) is decreasing.
Explanation:
To determine where the function f(x) = 4/(x^2 - 2x - 3) is decreasing, we need to examine the behavior of the function's derivative, f'(x), and identify the intervals where f'(x) is negative (indicating a decreasing function).
First, let's find the derivative of f(x):
f(x) = 4/(x^2 - 2x - 3)
To find f'(x), we'll use the quotient rule, which states that if you have a function of the form g(x)/h(x), the derivative is [g'(x)h(x) - g(x)h'(x)] / [h(x)]^2.
In our case, g(x) = 4 and h(x) = x^2 - 2x - 3. Therefore, we have:
g'(x) = 0 (since it's a constant)
h'(x) = 2x - 2 (derivative of the denominator)
Now, apply the quotient rule:
f'(x) = [0 * (x^2 - 2x - 3) - 4 * (2x - 2)] / (x^2 - 2x - 3)^2
f'(x) = (-8x + 8) / (x^2 - 2x - 3)^2
Now, we want to find where f'(x) is negative. This means that the numerator (-8x + 8) must be negative, while the denominator (x^2 - 2x - 3)^2 must be positive (since the square of a real number is always positive).
Numerator: -8x + 8 < 0
-8x < -8
x > 1
Denominator: Since it's squared, it's always positive.
So, for x > 1, the function f(x) = 4/(x^2 - 2x - 3) is decreasing