1. **State the problem:** Solve the inequality $$\frac{x}{x-2} + \frac{x}{x^2-4} \leq 0$$ using the graph of $$f(x) = \frac{x}{x-2} + \frac{x}{x^2-4}$$. 2. **Rewrite the function:** Note that $$x^2 - 4 = (x-2)(x+2)$$, so $$f(x) = \frac{x}{x-2} + \frac{x}{(x-2)(x+2)} = \frac{x(x+2)}{(x-2)(x+2)} + \frac{x}{(x-2)(x+2)} = \frac{x(x+2) + x}{(x-2)(x+2)} = \frac{x^2 + 2x + x}{(x-2)(x+2)} = \frac{x^2 + 3x}{(x-2)(x+2)}.$$ 3. **Simplify numerator:** $$x^2 + 3x = x(x+3).$$ 4. **Rewrite inequality:** $$\frac{x(x+3)}{(x-2)(x+2)} \leq 0.$$ 5. **Identify critical points:** The numerator zeros are at $$x=0$$ and $$x=-3$$. The denominator zeros (vertical asymptotes) are at $$x=2$$ and $$x=-2$$ (excluded from domain). 6. **Determine sign intervals:** The critical points divide the real line into intervals: $$(-\infty, -3), (-3, -2), (-2, 0), (0, 2), (2, \infty).$$ 7. **Test signs on each interval:** - For $$x < -3$$, pick $$x=-4$$: numerator $$x(x+3) = (-4)(-1) = 4 > 0$$, denominator $$(x-2)(x+2) = (-6)(-2) = 12 > 0$$, so $$f(x) > 0$$. - For $$-3 < x < -2$$, pick $$x=-2.5$$: numerator $$(-2.5)(0.5) = -1.25 < 0$$, denominator $$(-4.5)(-0.5) = 2.25 > 0$$, so $$f(x) < 0$$. - For $$-2 < x < 0$$, pick $$x=-1$$: numerator $$(-1)(2) = -2 < 0$$, denominator $$(-3)(1) = -3 < 0$$, so $$f(x) > 0$$. - For $$0 < x < 2$$, pick $$x=1$$: numerator $$(1)(4) = 4 > 0$$, denominator $$(-1)(3) = -3 < 0$$, so $$f(x) < 0$$. - For $$x > 2$$, pick $$x=3$$: numerator $$(3)(6) = 18 > 0$$, denominator $$(1)(5) = 5 > 0$$, so $$f(x) > 0$$. 8. **Include points where $$f(x) = 0$$:** Numerator zeroes at $$x=-3$$ and $$x=0$$, so $$f(x) = 0$$ there. 9. **Domain restrictions:** $$x \neq -2$$ and $$x \neq 2$$ because denominator zero. 10. **Solution to inequality $$f(x) \leq 0$$:** Intervals where $$f(x) \leq 0$$ are where $$f(x) < 0$$ or $$f(x) = 0$$: - $$-3 \leq x < -2$$ (includes $$-3$$ where $$f(x)=0$$, excludes $$-2$$) - $$0 \leq x < 2$$ (includes $$0$$ where $$f(x)=0$$, excludes $$2$$) **Final answer:** $$\boxed{-3 \leq x < -2, \quad 0 \leq x < 2}$$. This corresponds to option (d).