1. **State the problem:** Given the function $f(x) = \frac{x^4 + 1}{x^2}$, determine the intervals where the function is decreasing. 2. **Rewrite the function:** Simplify $f(x)$ by dividing each term in the numerator by $x^2$: $$f(x) = \frac{x^4}{x^2} + \frac{1}{x^2} = x^2 + x^{-2}$$ 3. **Find the derivative:** To analyze where $f(x)$ is increasing or decreasing, compute $f'(x)$: $$f'(x) = \frac{d}{dx}(x^2 + x^{-2}) = 2x - 2x^{-3} = 2x - \frac{2}{x^3}$$ 4. **Simplify derivative:** $$f'(x) = 2x - \frac{2}{x^3} = \frac{2x^4 - 2}{x^3} = \frac{2(x^4 - 1)}{x^3}$$ 5. **Critical points and domain:** - The derivative is undefined at $x = 0$, which is outside the domain since $f(x)$ is undefined there. - Set numerator to zero to find where $f'(x)=0$: $$x^4 - 1 = 0 \implies x^4 = 1 \implies x = \pm 1$$ 6. **Determine sign of $f'(x)$ on intervals defined by $-\infty, -1, 0, 1, \infty$:** - For $x < -1$: choose $x = -2$, $$f'(-2) = \frac{2((-2)^4 - 1)}{(-2)^3} = \frac{2(16 - 1)}{-8} = \frac{2 \times 15}{-8} = -\frac{30}{8} < 0$$ so $f'(x)<0$, function is decreasing. - For $-1 < x < 0$: choose $x = -\frac{1}{2}$, $$f'\left(-\frac{1}{2}\right) = \frac{2\left(\left(-\frac{1}{2}\right)^4 - 1\right)}{\left(-\frac{1}{2}\right)^3} = \frac{2\left(\frac{1}{16} - 1\right)}{-\frac{1}{8}} = \frac{2\left(-\frac{15}{16}\right)}{-\frac{1}{8}} = \frac{-\frac{30}{16}}{-\frac{1}{8}} = \frac{-1.875}{-0.125} = 15 > 0$$ so $f'(x)>0$, function is increasing. - For $0 < x < 1$: choose $x = \frac{1}{2}$, $$f'\left(\frac{1}{2}\right) = \frac{2\left(\left(\frac{1}{2}\right)^4 - 1\right)}{\left(\frac{1}{2}\right)^3} = \frac{2\left(\frac{1}{16} - 1\right)}{\frac{1}{8}} = \frac{2\left(-\frac{15}{16}\right)}{\frac{1}{8}} = \frac{-\frac{30}{16}}{\frac{1}{8}} = \frac{-1.875}{0.125} = -15 < 0$$ so $f'(x) < 0$, function is decreasing. - For $x > 1$: choose $x = 2$, $$f'(2) = \frac{2(16 - 1)}{8} = \frac{2 \times 15}{8} = \frac{30}{8} = 3.75 > 0$$ so $f'(x) > 0$, function is increasing. 7. **Conclusion on decreasing intervals:** The function is decreasing where $f'(x) < 0$, which is on intervals: $$(-\infty, -1) \quad \text{and} \quad (0, 1)$$ 8. **Answer choice:** This matches option (d). **Final answer:** The function $f(x) = \frac{x^4 + 1}{x^2}$ is decreasing on the intervals $(-\infty, -1)$ and $(0, 1)$.