1. **Problem Statement:** Determine on which intervals the function $f(x) = |9 - x^2|$ is decreasing. 2. **Understanding the function:** The function is the absolute value of $9 - x^2$. This means: - When $9 - x^2 \geq 0$, $f(x) = 9 - x^2$. - When $9 - x^2 < 0$, $f(x) = -(9 - x^2) = x^2 - 9$. 3. **Find critical points:** Solve $9 - x^2 = 0$ to find where the expression inside the absolute value changes sign. $$9 - x^2 = 0 \implies x^2 = 9 \implies x = \pm 3$$ 4. **Analyze intervals based on critical points:** - For $x < -3$, $9 - x^2 < 0$ so $f(x) = x^2 - 9$. - For $-3 \leq x \leq 3$, $9 - x^2 \geq 0$ so $f(x) = 9 - x^2$. - For $x > 3$, $9 - x^2 < 0$ so $f(x) = x^2 - 9$. 5. **Find derivative on each interval to determine increasing/decreasing behavior:** - For $x < -3$ and $x > 3$, $f(x) = x^2 - 9$, so $$f'(x) = 2x$$ - For $-3 < x < 3$, $f(x) = 9 - x^2$, so $$f'(x) = -2x$$ 6. **Determine where $f'(x) < 0$ (decreasing):** - For $x < -3$, $f'(x) = 2x$. Since $x < -3$, $2x < 0$, so $f$ is decreasing on $(-\infty, -3)$. - For $-3 < x < 3$, $f'(x) = -2x$. For $x$ in $(0,3)$, $-2x < 0$, so $f$ is decreasing on $(0,3)$. - For $-3 < x < 0$, $-2x > 0$, so $f$ is increasing there. - For $x > 3$, $f'(x) = 2x > 0$, so $f$ is increasing on $(3, \infty)$. 7. **Combine intervals where $f$ is decreasing:** $$(-\infty, -3) \cup (0, 3)$$ 8. **Answer:** The function $f(x) = |9 - x^2|$ is decreasing on the intervals $(-\infty, -3) \cup (0, 3)$, which corresponds to option A. **Final answer:** A) $] - \infty, -3 [ \cup ] 0, 3 [$