Question: The rational function r is given by r(x) = $\frac{x^3 + 4x^2 + 4x}{x^2 - 9}$. On what intervals of x is r(x) $\geq 0$ ?
A $x \geq 0$
B $-3 < x < 3$
C $-3 < x < -2$, $-2 < x < 0$, and $x > 3$ only
D $-3 < x \leq 0$ and $x > 3$
1. **State the problem:** We want to find the intervals where the rational function $$r(x) = \frac{x^3 + 4x^2 + 4x}{x^2 - 9}$$ is greater than or equal to zero, i.e., $$r(x) \geq 0.$$
2. **Factor numerator and denominator:**
- Numerator: $$x^3 + 4x^2 + 4x = x(x^2 + 4x + 4) = x(x+2)^2.$$
- Denominator: $$x^2 - 9 = (x-3)(x+3).$$
So, $$r(x) = \frac{x(x+2)^2}{(x-3)(x+3)}.$$
3. **Identify critical points:** These are points where numerator or denominator is zero:
- Numerator zeros: $x=0$, $x=-2$ (with multiplicity 2)
- Denominator zeros: $x=3$, $x=-3$ (vertical asymptotes, function undefined)
4. **Determine sign intervals:** The critical points divide the real line into intervals:
$$(-\infty, -3), (-3, -2), (-2, 0), (0, 3), (3, \infty).$$
5. **Test sign of $r(x)$ on each interval:**
- For $x < -3$, pick $x=-4$:
- Numerator: $x(x+2)^2 = (-4)(-2)^2 = (-4)(4) = -16 < 0$
- Denominator: $(x-3)(x+3) = (-7)(-1) = 7 > 0$
- $r(x) = \frac{negative}{positive} = negative$
- For $-3 < x < -2$, pick $x=-2.5$:
- Numerator: $(-2.5)((-0.5)^2) = (-2.5)(0.25) = -0.625 < 0$
- Denominator: $(-2.5-3)(-2.5+3) = (-5.5)(0.5) = -2.75 < 0$
- $r(x) = \frac{negative}{negative} = positive$
- For $-2 < x < 0$, pick $x=-1$:
- Numerator: $(-1)(1)^2 = -1 < 0$
- Denominator: $(-1-3)(-1+3) = (-4)(2) = -8 < 0$
- $r(x) = \frac{negative}{negative} = positive$
- For $0 < x < 3$, pick $x=1$:
- Numerator: $(1)(3)^2 = 1*9 = 9 > 0$
- Denominator: $(1-3)(1+3) = (-2)(4) = -8 < 0$
- $r(x) = \frac{positive}{negative} = negative$
- For $x > 3$, pick $x=4$:
- Numerator: $(4)(6)^2 = 4*36 = 144 > 0$
- Denominator: $(4-3)(4+3) = (1)(7) = 7 > 0$
- $r(x) = \frac{positive}{positive} = positive$
6. **Check values at critical points:**
- At $x=-3$ and $x=3$, denominator zero, so $r(x)$ undefined.
- At $x=-2$, numerator zero, so $r(-2) = 0$.
- At $x=0$, numerator zero, so $r(0) = 0$.
7. **Combine intervals where $r(x) \geq 0$:**
- From step 5, $r(x) > 0$ on $(-3, -2)$, $(-2, 0)$, and $(3, \infty)$.
- At $x=-2$ and $x=0$, $r(x) = 0$, so include these points.
Therefore, the solution is:
$$-3 < x \leq 0 \quad \text{and} \quad x > 3.$$
**Answer: Option D.**