1. **Problem Statement:** Analyze the function $$f(x) = \frac{x^2}{x^2 + 4}$$ for domain, range, first and second derivatives, intervals of increase/decrease, concavity, inflection points, and asymptotes. 2. **Domain:** The denominator $$x^2 + 4$$ is always positive since $$x^2 \geq 0$$ and 4 is positive. So, the domain is all real numbers: $$(-\infty, \infty)$$. 3. **Range:** Since $$x^2 \geq 0$$, $$f(x) = \frac{x^2}{x^2 + 4}$$ is always between 0 and 1. When $$x=0$$, $$f(0) = 0$$. As $$x \to \pm \infty$$, $$f(x) \to 1$$. So, the range is $$[0, 1)$$. 4. **First Derivative:** Use the quotient rule: $$f'(x) = \frac{(2x)(x^2 + 4) - x^2(2x)}{(x^2 + 4)^2} = \frac{2x(x^2 + 4) - 2x^3}{(x^2 + 4)^2}$$. Simplify numerator: $$2x^3 + 8x - 2x^3 = 8x$$. So, $$f'(x) = \frac{8x}{(x^2 + 4)^2}$$. 5. **Critical Points and Intervals of Increase/Decrease:** - Set $$f'(x) = 0$$: $$8x = 0 \Rightarrow x = 0$$. - For $$x < 0$$, $$f'(x) < 0$$ (function decreasing). - For $$x > 0$$, $$f'(x) > 0$$ (function increasing). 6. **Second Derivative:** Differentiate $$f'(x)$$ using quotient rule: $$f''(x) = \frac{8(x^2 + 4)^2 - 8x \cdot 2(x^2 + 4)(2x)}{(x^2 + 4)^4}$$. Simplify numerator: $$8(x^2 + 4)^2 - 32x^2(x^2 + 4) = 8(x^2 + 4)[(x^2 + 4) - 4x^2] = 8(x^2 + 4)(-3x^2 + 4)$$. So, $$f''(x) = \frac{8(x^2 + 4)(-3x^2 + 4)}{(x^2 + 4)^4} = \frac{8(-3x^2 + 4)}{(x^2 + 4)^3}$$. 7. **Concavity and Inflection Points:** - Set $$f''(x) = 0$$: $$-3x^2 + 4 = 0 \Rightarrow x^2 = \frac{4}{3} \Rightarrow x = \pm \frac{2}{\sqrt{3}}$$. - For $$|x| < \frac{2}{\sqrt{3}}$$, $$f''(x) > 0$$ (concave up). - For $$|x| > \frac{2}{\sqrt{3}}$$, $$f''(x) < 0$$ (concave down). 8. **Asymptotes:** - Horizontal asymptote as $$x \to \pm \infty$$ is $$y = 1$$ since $$f(x) \to 1$$. - No vertical asymptotes since denominator never zero. **Final summary:** - Domain: $$(-\infty, \infty)$$ - Range: $$[0, 1)$$ - Increasing on $$(0, \infty)$$, decreasing on $$(-\infty, 0)$$ - Concave up on $$\left(-\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\right)$$, concave down elsewhere - Inflection points at $$x = \pm \frac{2}{\sqrt{3}}$$ - Horizontal asymptote at $$y=1$$