1. **Problem Statement:** Analyze the function $$f(x) = \frac{x^2}{x^2 - 25}$$ for intervals of increase and decrease, local maxima and minima, intervals of concavity, and inflection points. 2. **Domain:** The denominator cannot be zero, so $$x^2 - 25 \neq 0 \Rightarrow x \neq \pm 5$$. The domain is $$(-\infty, -5) \cup (-5, 5) \cup (5, \infty)$$. 3. **First Derivative (for increasing/decreasing and local extrema):** Use quotient rule: $$f'(x) = \frac{(2x)(x^2 - 25) - x^2(2x)}{(x^2 - 25)^2} = \frac{2x(x^2 - 25) - 2x^3}{(x^2 - 25)^2}$$ Simplify numerator: $$2x(x^2 - 25) - 2x^3 = 2x^3 - 50x - 2x^3 = -50x$$ So, $$f'(x) = \frac{-50x}{(x^2 - 25)^2}$$ 4. **Critical Points:** Set numerator zero: $$-50x = 0 \Rightarrow x = 0$$ Check if in domain: yes. 5. **Sign of $$f'(x)$$:** - For $$x < 0$$, numerator $$-50x > 0$$ (since $$x$$ negative), denominator always positive (square), so $$f'(x) > 0$$. - For $$x > 0$$, numerator $$-50x < 0$$, denominator positive, so $$f'(x) < 0$$. 6. **Intervals of Increase/Decrease:** - Increasing on $$(-\infty, -5)$$ and $$(-5, 0)$$. - Decreasing on $$(0, 5)$$ and $$(5, \infty)$$. 7. **Local Extrema:** At $$x=0$$, $$f'(x)$$ changes from positive to negative, so local maximum. Calculate $$f(0) = \frac{0}{0 - 25} = 0$$. 8. **Second Derivative (for concavity and inflection points):** Rewrite $$f'(x) = -50x (x^2 - 25)^{-2}$$. Use product and chain rules: $$f''(x) = -50 \left[(x^2 - 25)^{-2} + x \cdot (-2)(x^2 - 25)^{-3} \cdot 2x \right]$$ Simplify: $$f''(x) = -50 \left[(x^2 - 25)^{-2} - 4x^2 (x^2 - 25)^{-3} \right] = -50 (x^2 - 25)^{-3} \left[(x^2 - 25) - 4x^2 \right]$$ $$= -50 (x^2 - 25)^{-3} (-3x^2 - 25) = 50 (x^2 - 25)^{-3} (3x^2 + 25)$$ 9. **Sign of $$f''(x)$$:** - Numerator $$50(3x^2 + 25) > 0$$ for all real $$x$$. - Denominator $$ (x^2 - 25)^3 $$ changes sign at $$x = \pm 5$$. - For $$|x| > 5$$, $$x^2 - 25 > 0$$, so denominator positive, $$f''(x) > 0$$ (concave up). - For $$|x| < 5$$, $$x^2 - 25 < 0$$, denominator negative, $$f''(x) < 0$$ (concave down). 10. **Inflection Points:** Points where concavity changes are at $$x = -5$$ and $$x = 5$$, but these are not in the domain (vertical asymptotes), so no inflection points. **Final answers:** - Increasing on $$(-\infty, -5) \cup (-5, 0)$$. - Decreasing on $$(0, 5) \cup (5, \infty)$$. - Local maximum at $$x=0$$ with value $$0$$. - Concave down on $$(-5, 5)$$. - Concave up on $$(-\infty, -5) \cup (5, \infty)$$. - No inflection points in the domain.