the graph of f(x) = 4 / (x^2 - 2x - 3) is shown. for which values of x is f(x) decreasing? (1, 3) ∪ (3, ∞)…

Explanation:

Step1: Recall decreasing - function property

A function $y = f(x)$ is decreasing when $f'(x)<0$. First, factor the denominator of $f(x)=\frac{4}{x^{2}-2x - 3}=\frac{4}{(x - 3)(x+1)}$. The domain of $f(x)$ is $x\neq - 1,x\neq3$ since the denominator cannot be zero.

Step2: Use the quotient - rule to find the derivative

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 4$, $u'=0$, $v=x^{2}-2x - 3$, and $v'=2x - 2$. So, $f'(x)=\frac{0\times(x^{2}-2x - 3)-4\times(2x - 2)}{(x^{2}-2x - 3)^{2}}=\frac{-8(x - 1)}{(x^{2}-2x - 3)^{2}}$.

Step3: Find where $f'(x)<0$

The denominator $(x^{2}-2x - 3)^{2}=(x - 3)^{2}(x + 1)^{2}>0$ for all $x\neq - 1,x\neq3$. We need to solve the inequality $\frac{-8(x - 1)}{(x^{2}-2x - 3)^{2}}<0$. Since the denominator is always positive for $x\neq - 1,x\neq3$, we focus on the numerator. Solving $-8(x - 1)<0$ gives $x>1$. Combining with the domain $x\neq - 1,x\neq3$, we get that $f(x)$ is decreasing on the intervals $(1,3)\cup(3,\infty)$.

Answer:

$(1,3)\cup(3,\infty)$