1. Stating the problem: Solve the inequality $$\frac{x^2 - 4}{x^{-x} - 2} > 0$$. 2. Analyze the numerator and denominator separately. - Numerator: $$x^2 - 4 = (x-2)(x+2)$$. - Denominator: $$x^{-x} - 2$$ is a complicated expression involving an exponent with variable base and exponent. 3. Since the denominator is complex, let's clarify if the expression is $$\frac{x^2 - 4}{x^{-x} - 2} > 0$$ or if there is a typo. Assuming the denominator is $$x^{-x} - 2$$, this is not a standard algebraic expression and requires numerical or graphical methods. 4. However, if the denominator is meant to be $$x^{-x} - 2$$, note that $$x^{-x} = \frac{1}{x^x}$$, which is defined only for positive $$x$$. 5. Domain: $$x > 0$$ to have $$x^x$$ defined. 6. Rewrite the inequality: $$\frac{(x-2)(x+2)}{\frac{1}{x^x} - 2} > 0$$ 7. Simplify denominator: $$\frac{1}{x^x} - 2 = \frac{1 - 2x^x}{x^x}$$ 8. So the inequality becomes: $$\frac{(x-2)(x+2)}{\frac{1 - 2x^x}{x^x}} > 0 \implies (x-2)(x+2) \cdot \frac{x^x}{1 - 2x^x} > 0$$ 9. Since $$x^x > 0$$ for $$x > 0$$, the sign depends on: $$ (x-2)(x+2) \cdot \frac{1}{1 - 2x^x} > 0$$ 10. Critical points from numerator: $$x = -2, 2$$ (only $$x=2$$ in domain $$x>0$$). 11. Critical points from denominator: solve $$1 - 2x^x = 0 \Rightarrow x^x = \frac{1}{2}$$. 12. Numerically approximate $$x$$ such that $$x^x = 0.5$$. - At $$x=0.5$$, $$0.5^{0.5} = \sqrt{0.5} \approx 0.707 > 0.5$$. - At $$x=0.3$$, $$0.3^{0.3} \approx 0.696 > 0.5$$. - At $$x=0.1$$, $$0.1^{0.1} \approx 0.794 > 0.5$$. Actually, $$x^x$$ decreases from 1 at $$x=1$$ to 0 at 0, but the minimum is at $$x = e^{-1} \approx 0.3679$$. Check at $$x=0.2$$: $$0.2^{0.2} \approx 0.724 > 0.5$$. Check at $$x=0.05$$: $$0.05^{0.05} \approx 0.860 > 0.5$$. So $$x^x > 0.5$$ for all $$x > 0$$. Therefore, $$1 - 2x^x < 0$$ for all $$x > 0$$ because $$2x^x > 1$$. 13. So denominator $$1 - 2x^x < 0$$ for all $$x > 0$$. 14. Sign analysis: - Numerator factors: $$x-2$$ and $$x+2$$. - For $$x > 0$$, $$x+2 > 0$$ always. - So numerator sign depends on $$x-2$$. 15. Denominator is negative for all $$x > 0$$. 16. Inequality: $$\frac{(x-2)(x+2)}{x^{-x} - 2} > 0 \implies (x-2)(x+2) \cdot \frac{1}{1 - 2x^x} > 0$$ Since denominator is negative, the inequality holds when numerator is negative: $$(x-2)(x+2) < 0$$ for $$x > 0$$. 17. For $$x > 0$$, $$x+2 > 0$$, so sign depends on $$x-2$$. $$(x-2) < 0 \implies x < 2$$. 18. Domain is $$x > 0$$, so solution is: $$0 < x < 2$$. 19. Check endpoints: - At $$x=2$$ numerator zero, fraction zero, not greater than zero. - At $$x=0$$ denominator undefined. 20. Final solution: $$\boxed{0 < x < 2}$$