1. **Problem Statement:** We are given the function $$f(x) = \frac{a}{x^2 + b^2}$$ where $a, b > 0$. We need to find the critical points and possible points of inflection in terms of $a$ and $b$. 2. **Recall:** - Critical points occur where the first derivative $f'(x) = 0$ or is undefined. - Points of inflection occur where the second derivative $f''(x) = 0$ and the concavity changes. 3. **Find the first derivative:** Using the quotient rule or rewriting $f(x) = a(x^2 + b^2)^{-1}$, $$f'(x) = a \cdot (-1)(x^2 + b^2)^{-2} \cdot 2x = -\frac{2ax}{(x^2 + b^2)^2}$$ 4. **Find critical points:** Set $f'(x) = 0$: $$-\frac{2ax}{(x^2 + b^2)^2} = 0 \implies 2ax = 0$$ Since $a > 0$, this implies $$x = 0$$ 5. **Find the second derivative:** Differentiate $f'(x)$: $$f''(x) = \frac{d}{dx} \left(-\frac{2ax}{(x^2 + b^2)^2}\right)$$ Use the product and chain rules: $$f''(x) = -2a \cdot \frac{(x^2 + b^2)^2 - x \cdot 2 (x^2 + b^2) \cdot 2x}{(x^2 + b^2)^4}$$ Simplify numerator: $$ (x^2 + b^2)^2 - 4x^2 (x^2 + b^2) = (x^2 + b^2)(x^2 + b^2 - 4x^2) = (x^2 + b^2)(-3x^2 + b^2)$$ So, $$f''(x) = -2a \cdot \frac{(x^2 + b^2)(-3x^2 + b^2)}{(x^2 + b^2)^4} = -2a \cdot \frac{-3x^2 + b^2}{(x^2 + b^2)^3} = \frac{2a(3x^2 - b^2)}{(x^2 + b^2)^3}$$ 6. **Find points of inflection:** Set $f''(x) = 0$: $$\frac{2a(3x^2 - b^2)}{(x^2 + b^2)^3} = 0 \implies 3x^2 - b^2 = 0$$ Solve for $x$: $$3x^2 = b^2 \implies x^2 = \frac{b^2}{3} \implies x = \pm \frac{b}{\sqrt{3}}$$ 7. **Summary:** - Critical point at $$x = 0$$ - Possible points of inflection at $$x = -\frac{b}{\sqrt{3}}$$ (smaller value) and $$x = \frac{b}{\sqrt{3}}$$ (larger value) These points depend only on $b$ and not on $a$ (except $a > 0$ ensures the function shape). **Final answer:** - Critical point: $$x = 0$$ - Points of inflection: $$x = -\frac{b}{\sqrt{3}}, \quad x = \frac{b}{\sqrt{3}}$$