1. **State the problem:** We are given the derivative of a function $f'(x)$ and want to find the original function $f(x)$ and analyze its critical points. 2. **Given:** $$f'(x) = \frac{(-3x^2)(x^2-4) - (-x^3)(2x)}{(x^2-4)^2} = \frac{x^4 + 12x^2}{(x^2-4)^2}$$ 3. **Simplify the numerator:** $$(-3x^2)(x^2-4) - (-x^3)(2x) = (-3x^4 + 12x^2) + 2x^4 = x^4 + 12x^2$$ 4. **Rewrite $f'(x)$:** $$f'(x) = \frac{x^4 + 12x^2}{(x^2-4)^2}$$ 5. **Find $f(x)$ by integrating $f'(x)$:** Notice the numerator can be factored: $$x^4 + 12x^2 = x^2(x^2 + 12)$$ So, $$f(x) = \frac{x^2(x^2 + 12)}{(x^2 - 4)^2}$$ 6. **Identify critical points:** Critical points occur where $f'(x) = 0$ or is undefined. - Numerator zero: $x^2(x^2 + 12) = 0 \Rightarrow x = 0$ (since $x^2 + 12 > 0$ always) - Denominator zero: $x^2 - 4 = 0 \Rightarrow x = \pm 2$ (undefined points) 7. **Domain restrictions:** $$x \neq \pm 2$$ 8. **Sign analysis of $f'(x)$:** Intervals: $(-\infty, -2), (-2, 0), (0, 2), (2, \infty)$ - For $x < -2$, numerator $> 0$, denominator $> 0$, so $f'(x) > 0$ - For $-2 < x < 0$, numerator $> 0$, denominator $> 0$, so $f'(x) > 0$ - For $0 < x < 2$, numerator $> 0$, denominator $> 0$, so $f'(x) > 0$ - For $x > 2$, numerator $> 0$, denominator $> 0$, so $f'(x) > 0$ Since $f'(x) > 0$ everywhere in the domain, $f(x)$ is increasing on each interval. **Final answer:** $$f(x) = \frac{x^2(x^2 + 12)}{(x^2 - 4)^2}, \quad x \neq \pm 2$$ Critical point at $x=0$ where $f'(0) = 0$, but function is increasing on both sides.