1. **Problem Statement:** Find the extreme values (absolute and local) of the function $$y = \frac{x^2}{x+2}$$ over its natural domain. 2. **Domain:** The function is undefined where the denominator is zero, so $$x+2 \neq 0 \Rightarrow x \neq -2$$. The domain is $$(-\infty, -2) \cup (-2, \infty)$$. 3. **Find the derivative:** Use the quotient rule: $$y = \frac{f(x)}{g(x)} = \frac{x^2}{x+2}$$ $$f'(x) = 2x, \quad g'(x) = 1$$ $$y' = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} = \frac{2x(x+2) - x^2(1)}{(x+2)^2} = \frac{2x^2 + 4x - x^2}{(x+2)^2} = \frac{x^2 + 4x}{(x+2)^2}$$ 4. **Find critical points:** Set numerator equal to zero (denominator squared is never zero except at excluded point): $$x^2 + 4x = 0$$ $$x(x + 4) = 0$$ So, $$x = 0$$ or $$x = -4$$. 5. **Check if critical points are in domain:** Both $$0$$ and $$-4$$ are in the domain. 6. **Evaluate function at critical points:** $$y(0) = \frac{0^2}{0+2} = 0$$ $$y(-4) = \frac{(-4)^2}{-4+2} = \frac{16}{-2} = -8$$ 7. **Check behavior near domain endpoints:** - As $$x \to -2^-$$, denominator $$\to 0^-$$, numerator $$\to 4$$, so $$y \to -\infty$$. - As $$x \to -2^+$$, denominator $$\to 0^+$$, numerator $$\to 4$$, so $$y \to +\infty$$. - As $$x \to \pm \infty$$, $$y \approx \frac{x^2}{x} = x \to \pm \infty$$. 8. **Determine nature of critical points using first derivative test:** - For $$x = -4$$: - Pick $$x = -5$$: $$y' = \frac{25 - 20}{( -5 + 2)^2} = \frac{5}{9} > 0$$ (increasing) - Pick $$x = -3$$: $$y' = \frac{9 - 12}{( -3 + 2)^2} = \frac{-3}{1} < 0$$ (decreasing) So, $$x = -4$$ is a local maximum. - For $$x = 0$$: - Pick $$x = -1$$: $$y' = \frac{1 - 4}{( -1 + 2)^2} = \frac{-3}{1} < 0$$ (decreasing) - Pick $$x = 1$$: $$y' = \frac{1 + 4}{(1 + 2)^2} = \frac{5}{9} > 0$$ (increasing) So, $$x = 0$$ is a local minimum. 9. **Summary of extrema:** - Local maximum at $$x = -4$$ with $$y = -8$$. - Local minimum at $$x = 0$$ with $$y = 0$$. 10. **Absolute extrema:** - No absolute maximum because $$y \to +\infty$$ as $$x \to -2^+$$. - No absolute minimum because $$y \to -\infty$$ as $$x \to -2^-$$. Final answer: - Local maximum: $$(-4, -8)$$ - Local minimum: $$(0, 0)$$