Question

the graph of ( f(x) = \frac{4}{x^2 - 2x - 3} ) is shown. for which values of ( x ) is ( f(x) ) decreasing?
options:
( (1, 3) cup (3, infty) )
( (-1, 3) cup (3, infty) )
( (3, infty) )
( (-1, 3) )

Explanation:

Step1: Analyze the graph's behavior

The function \( f(x)=\frac{4}{x^2 - 2x - 3}\) (factoring denominator: \(x^2 - 2x - 3=(x - 3)(x + 1)\), so vertical asymptotes at \(x=-1\) and \(x = 3\)). From the graph, we observe the intervals of decrease. The middle part (between the asymptotes \(x=-1\) and \(x = 3\)) of the graph (the downward - opening curve) is decreasing? Wait, no, wait. Wait, the graph: the left part (left of \(x=-1\)) is increasing, the middle part (between \(x=-1\) and \(x = 3\)): let's see the middle curve (the one below the x - axis) has a maximum, so it first increases then decreases? Wait, no, the question is about where \(f(x)\) is decreasing. Wait, looking at the options, the correct interval for decrease: let's check the graph. The middle region (between \(x=-1\) and \(x = 3\))? Wait, no, wait the options: the correct answer is \((-1,3)\)? Wait, no, wait the function's derivative: \(f^\prime(x)=4\times\frac{-(2x - 2)}{(x^2 - 2x - 3)^2}=\frac{-8(x - 1)}{(x^2 - 2x - 3)^2}\). The critical point at \(x = 1\). But from the graph, the middle part (between \(x=-1\) and \(x = 3\)): the lower curve (the one with the maximum) - wait, maybe the graph shows that between \(x=-1\) and \(x = 3\), the function (the lower part) is decreasing? Wait, the options: the fourth option is \((-1,3)\). Wait, let's check the graph again. The left asymptote is at \(x=-1\), right at \(x = 3\). The middle curve (the one below the x - axis) has a peak, so it increases from \(x=-1\) to \(x = 1\) and then decreases from \(x = 1\) to \(x = 3\)? No, that can't be. Wait, maybe I misread. Wait the function is \(f(x)=\frac{4}{x^2-2x - 3}\). The denominator is \(x^2-2x - 3=(x - 3)(x + 1)\). So the domain is \(x eq - 1,x eq3\). The graph: left of \(x=-1\): as \(x\to-\infty\), \(f(x)\to0^+\); as \(x\to - 1^-\), \(f(x)\to+\infty\) (so increasing). Between \(x=-1\) and \(x = 3\): the function is negative (since denominator is negative), and as \(x\) increases from \(-1\) to \(3\), the denominator \(x^2-2x - 3\) goes from \(0^-\) to \(0^-\) (wait, at \(x = 1\), denominator is \(1 - 2 - 3=-4\)). Wait, the graph shows a downward - opening parabola - like curve between \(x=-1\) and \(x = 3\) (the lower part) and two hyperbolic - like curves on the left (of \(x=-1\)) and right (of \(x = 3\)). The right - hand curve (for \(x>3\)): as \(x\to+\infty\), \(f(x)\to0^+\); as \(x\to3^+\), \(f(x)\to+\infty\) (so decreasing). The left - hand curve (for \(x<-1\)): as \(x\to-\infty\), \(f(x)\to0^+\); as \(x\to - 1^-\), \(f(x)\to+\infty\) (so increasing). The middle curve (for \(-1<x<3\)): let's see the y - values. At \(x = 0\), \(f(0)=\frac{4}{-3}\approx - 1.33\); at \(x = 1\), \(f(1)=\frac{4}{1 - 2 - 3}=\frac{4}{-4}=-1\); at \(x = 2\), \(f(2)=\frac{4}{4 - 4 - 3}=\frac{4}{-3}\approx - 1.33\). Wait, so from \(x=-1\) to \(x = 1\), the function increases (from \(-\infty\) to \(-1\)) and from \(x = 1\) to \(x = 3\), it decreases (from \(-1\) to \(-\infty\))? But the options: the fourth option is \((-1,3)\). Wait, maybe the graph is interpreted as the middle region (between \(-1\) and \(3\)) is where the function is decreasing? Wait, the options: let's check the options again. The options are:

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

2. \((-1,3)\cup(3,\infty)\)

3. \((3,\infty)\)

4. \((-1,3)\)

Wait, maybe the graph shows that in the interval \((-1,3)\), the function (the lower curve) is decreasing? Wait, perhaps the key is to look at the graph's slope. The middle curve (between \(x=-1\) and \(x = 3\)) has a negative slope (decreasing) overall? Or maybe the question is about the function's decreasing interval. Let's re - express the function: \(f(x)=\frac{4}{(x - 1)^2-4}\). The denominator is a parabola opening upwards with vertex at \((1,-4)\). So the function \(f(x)\) has a maximum at \(x = 1\) (since the denominator is minimized at \(x = 1\), so \(f(x)\) is maximized there). So for \(x\in(-1,1)\), as \(x\) increases, the denominator \((x - 1)^2-4\) decreases (since \((x - 1)^2\) decreases as \(x\) approaches \(1\) from the left), so \(f(x)=\frac{4}{\text{denominator}}\) (denominator is negative) increases (because denominator is more negative, so the fraction is less negative, i.e., increases). For \(x\in(1,3)\), as \(x\) increases, the denominator \((x - 1)^2-4\) increases (since \((x - 1)^2\) increases as \(x\) moves away from \(1\) to the right), so \(f(x)=\frac{4}{\text{denominator}}\) (denominator is negative) decreases (because denominator is less negative, so the fraction is more negative, i.e., decreases). But the left - hand side (\(x<-1\)): as \(x\) increases towards \(-1\), denominator \((x - 1)^2-4\) approaches \((-2)^2-4 = 0\) from the positive side? Wait no, at \(x=-2\), denominator is \((-3)^2-4 = 5\), so \(f(-2)=\frac{4}{5}\). At \(x=-1\), denominator is \(0\). So for \(x<-1\), as \(x\) increases, denominator \((x - 1)^2-4\) decreases (from \(+\infty\) to \(0^+\)), so \(f(x)=\frac{4}{\text{denominator}}\) decreases (from \(0^+\) to \(+\infty\))? Wait, no, if denominator decreases (gets smaller positive), then \(f(x)\) increases (since \(4\) divided by a smaller positive number is larger). So \(x<-1\): \(f(x)\) is increasing. For \(x>3\): denominator \((x - 1)^2-4\) is positive and increasing as \(x\) increases, so \(f(x)=\frac{4}{\text{denominator}}\) decreases (since denominator increases, fraction decreases). So the decreasing intervals are \((1,3)\) and \((3,\infty)\)? But that's option 1. Wait, but the graph: the right - hand curve (for \(x>3\)) is decreasing (as \(x\) increases, \(y\) decreases towards \(0\)). The middle curve: from \(x = 1\) to \(x = 3\), \(y\) decreases (from \(-1\) to \(-\infty\)). The left - hand curve: increasing. So the decreasing intervals are \((1,3)\) and \((3,\infty)\), which is option 1: \((1,3)\cup(3,\infty)\). Wait, but maybe I made a mistake. Wait the original function is \(f(x)=\frac{4}{x^2-2x - 3}\). Let's find the derivative:

\(f^\prime(x)=4\times\frac{-(2x - 2)}{(x^2 - 2x - 3)^2}=\frac{-8(x - 1)}{(x^2 - 2x - 3)^2}\)

The denominator \((x^2 - 2x - 3)^2=(x - 3)^2(x + 1)^2\) is always positive except at \(x=-1\) and \(x = 3\) (where it's zero). So the sign of \(f^\prime(x)\) depends on \(-8(x - 1)\).

- When \(x< - 1\): \(x - 1<0\), so \(-8(x - 1)>0\), so \(f^\prime(x)>0\) (function increasing).

- When \(-1<x<1\): \(x - 1<0\), so \(-8(x - 1)>0\), so \(f^\prime(x)>0\) (function increasing).

- When \(1<x<3\): \(x - 1>0\), so \(-8(x - 1)<0\), so \(f^\prime(x)<0\) (function decreasing).

- When \(x>3\): \(x - 1>0\), so \(-8(x - 1)<0\), so \(f^\prime(x)<0\) (function decreasing).

So the function is decreasing on \((1,3)\) and \((3,\infty)\), which is the first option.

Step2: Match with options

The intervals where \(f(x)\) is decreasing are \((1,3)\) (since at \(x = 3\) the function is undefined) and \((3,\infty)\), so the interval is \((1,3)\cup(3,\infty)\).

Answer:

\((1, 3)\cup(3, \infty)\) (corresponding to the first option)