Question

consider the function $f(x) = \frac{9 - x^2}{x^2 - 4}$. for which intervals is $f(x)$ positive? check all that apply. $(-\infty, -3)$ $(-3, -2)$ $(-2, 2)$ $(2, 3)$ $(\infty, 3)$

Explanation:

Step1: Factor the function

First, factor the numerator and the denominator. The numerator \(9 - x^2\) is a difference of squares, so it factors as \((3 - x)(3 + x)\). The denominator \(x^2 - 4\) is also a difference of squares, factoring as \((x - 2)(x + 2)\). So the function becomes:

$$ f(x)=\frac{(3 - x)(3 + x)}{(x - 2)(x + 2)} $$

We can also rewrite the numerator as \(-(x - 3)(x + 3)\) to make it easier to analyze the sign, so:

$$ f(x)=\frac{-(x - 3)(x + 3)}{(x - 2)(x + 2)} $$

Step2: Find critical points

The critical points are where the numerator or the denominator is zero. The numerator is zero when \(x = 3\) or \(x=-3\), and the denominator is zero when \(x = 2\) or \(x=-2\). These points divide the real number line into the intervals: \((-\infty, -3)\), \((-3, -2)\), \((-2, 2)\), \((2, 3)\), and \((3, \infty)\).

Step3: Test intervals for sign

• Interval \((-\infty, -3)\): Let's pick a test point, say \(x=-4\). Plug into the function:

Numerator: \(-(-4 - 3)(-4 + 3)=-( -7)(-1)= -7\) (negative)
Denominator: \((-4 - 2)(-4 + 2)=(-6)(-2)=12\) (positive)
So \(f(-4)=\frac{-7}{12}\) (negative). So this interval is negative.

• Interval \((-3, -2)\): Pick \(x=-2.5\).

Numerator: \(-(-2.5 - 3)(-2.5 + 3)=-( -5.5)(0.5)=2.75\) (positive)
Denominator: \((-2.5 - 2)(-2.5 + 2)=(-4.5)(-0.5)=2.25\) (positive)
So \(f(-2.5)=\frac{2.75}{2.25}\) (positive). So this interval is positive.

• Interval \((-2, 2)\): Pick \(x = 0\).

Numerator: \(-(0 - 3)(0 + 3)=-(-3)(3)=9\) (positive)
Denominator: \((0 - 2)(0 + 2)=(-2)(2)=-4\) (negative)
So \(f(0)=\frac{9}{-4}\) (negative). So this interval is negative.

• Interval \((2, 3)\): Pick \(x = 2.5\).

Numerator: \(-(2.5 - 3)(2.5 + 3)=-(-0.5)(5.5)=2.75\) (positive)
Denominator: \((2.5 - 2)(2.5 + 2)=(0.5)(4.5)=2.25\) (positive)
So \(f(2.5)=\frac{2.75}{2.25}\) (positive). So this interval is positive.

• Interval \((3, \infty)\): Pick \(x = 4\).

Numerator: \(-(4 - 3)(4 + 3)=-(1)(7)= -7\) (negative)
Denominator: \((4 - 2)(4 + 2)=(2)(6)=12\) (positive)
So \(f(4)=\frac{-7}{12}\) (negative).

Wait, but the last interval in the options is \((3, \infty)\)? Wait, the last option is \((3, \infty)\)? Wait, the original options have \((-\infty, -3)\), \((-3, -2)\), \((-2, 2)\), \((2, 3)\), and \((3, \infty)\)? Wait, the last option in the user's question is \((3, \infty)\)? Wait, the user wrote \((\infty, 3)\) which is probably a typo, should be \((3, \infty)\). But let's check the options again. The options are \((-\infty, -3)\), \((-3, -2)\), \((-2, 2)\), \((2, 3)\), \((3, \infty)\) (assuming the last one is a typo).

Wait, in our analysis, \((-3, -2)\) and \((2, 3)\) are positive. Let's re - check the \((2, 3)\) interval. For \(x = 2.5\), numerator: \(-(2.5 - 3)(2.5 + 3)=-(-0.5)(5.5)=2.75\) (positive), denominator: \((2.5 - 2)(2.5 + 2)=(0.5)(4.5)=2.25\) (positive), so positive. For \((-3, -2)\), \(x=-2.5\), numerator positive, denominator positive, so positive.

Wait, let's re - check the first interval \((-\infty, -3)\). For \(x=-4\), numerator: \(-(-4 - 3)(-4 + 3)=-( -7)(-1)= -7\) (negative), denominator: \((-4 - 2)(-4 + 2)=(-6)(-2)=12\) (positive), so negative. \((-2, 2)\): \(x = 0\), numerator: \(-(0 - 3)(0 + 3)=9\) (positive), denominator: \((0 - 2)(0 + 2)=-4\) (negative), so negative. \((3, \infty)\): \(x = 4\), numerator negative, denominator positive, so negative.

So the intervals where \(f(x)\) is positive are \((-3, -2)\) and \((2, 3)\).

Answer:

\((-3, -2)\), \((2, 3)\) (corresponding to the options: \((-3, -2)\) and \((2, 3)\))