To determine the intervals where the function [tex]\( f(x) = \frac{4}{x^2 - 2x - 3} \)[/tex] is decreasing, we should analyze the behavior of its derivative [tex]\( f'(x) \)[/tex].
1. First, we identify the function and its components:
[tex]\[
f(x) = \frac{4}{x^2 - 2x - 3}
\][/tex]
2. Next, we recognize that the derivative of [tex]\( f(x) \)[/tex] can be found using the quotient rule:
[tex]\[
f'(x) = \frac{d}{dx} \left( \frac{4}{x^2 - 2x - 3} \right)
\][/tex]
3. Let [tex]\( g(x) = 4 \)[/tex] and [tex]\( h(x) = x^2 - 2x - 3 \)[/tex]. The quotient rule is given by:
[tex]\[
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}
\][/tex]
Here, [tex]\( g'(x) = 0 \)[/tex] because [tex]\( g(x) \)[/tex] is a constant.
4. So, we have:
[tex]\[
f'(x) = \frac{0 \cdot (x^2 - 2x - 3) - 4 \cdot (2x - 2)}{(x^2 - 2x - 3)^2} = \frac{-4(2x - 2)}{(x^2 - 2x - 3)^2}
\][/tex]
[tex]\[
f'(x) = \frac{-8x + 8}{(x^2 - 2x - 3)^2}
\][/tex]
[tex]\[
f'(x) = \frac{8(1 - x)}{(x^2 - 2x - 3)^2}
\][/tex]
5. To determine where [tex]\( f(x) \)[/tex] is decreasing, we need to find where [tex]\( f'(x) < 0 \)[/tex]. The critical points occur where the numerator is zero:
[tex]\[
8(1 - x) = 0 \implies 1 - x = 0 \implies x = 1
\][/tex]
6. We also need to consider where the denominator [tex]\( (x^2 - 2x - 3)^2 \)[/tex] is zero, as these points are not in the domain of [tex]\( f(x) \)[/tex]. The denominator can be factored:
[tex]\[
x^2 - 2x - 3 = (x - 3)(x + 1)
\][/tex]
So, the zeroes of the denominator are at:
[tex]\[
x = -1 \quad \text{and} \quad x = 3
\][/tex]
7. We now have critical points and points of discontinuity at [tex]\( x = -1 \)[/tex], [tex]\( x = 1 \)[/tex], and [tex]\( x = 3 \)[/tex].
8. To figure out on which intervals [tex]\( f'(x) \)[/tex] is negative, we test the sign of [tex]\( f'(x) \)[/tex] in the intervals [tex]\((-\infty, -1)\)[/tex], [tex]\((-1, 1)\)[/tex], [tex]\((1, 3)\)[/tex], and [tex]\((3, \infty)\)[/tex]:
- For [tex]\( x \)[/tex] in [tex]\((-\infty, -1)\)[/tex], pick, for example, [tex]\( x = -2 \)[/tex]:
[tex]\[
f'(-2) = \frac{8(1 - (-2))}{((-2)^2 - 2(-2) - 3)^2} = \frac{8(1 + 2)}{(4 + 4 - 3)^2} = \frac{24}{5^2} = \frac{24}{25} > 0
\][/tex]
Hence, [tex]\( f(x) \)[/tex] is increasing on [tex]\((-\infty, -1)\)[/tex].
- For [tex]\( x \)[/tex] in [tex]\((-1, 1)\)[/tex], pick, for example, [tex]\( x = 0 \)[/tex]:
[tex]\[
f'(0) = \frac{8(1 - 0)}{(0 - 2(0) - 3)^2} = \frac{8 \cdot 1}{(-3)^2} = \frac{8}{9} < 0
\][/tex]
Hence, [tex]\( f(x) \)[/tex] is decreasing on [tex]\((-1, 1)\)[/tex].
- For [tex]\( x \)[/tex] in [tex]\((1, 3)\)[/tex], pick, for example, [tex]\( x = 2 \)[/tex]:
[tex]\[
f'(2) = \frac{8(1 - 2)}{(4 - 4 - 3)^2} = \frac{8(-1)}{(-3)^2} = \frac{-8}{9} < 0
\][/tex]
Hence, [tex]\( f(x) \)[/tex] is decreasing on [tex]\((1, 3)\)[/tex].
- For [tex]\( x \)[/tex] in [tex]\((3, \infty)\)[/tex], pick, for example, [tex]\( x = 4 \)[/tex]:
[tex]\[
f'(4) = \frac{8(1 - 4)}{(16 - 8 - 3)^2} = \frac{8(-3)}{5^2} = \frac{-24}{25} > 0
\][/tex]
Hence, [tex]\( f(x) \)[/tex] is increasing on [tex]\((3, \infty)\)[/tex].
So, the intervals where [tex]\( f(x) \)[/tex] is decreasing are:
[tex]\[
(-1, 1) \cup (1, 3) = (-1, 3)
\][/tex]
Therefore, the correct option is:
[tex]\[
(-1, 3)
\][/tex]
