1. **State the problem:** Given the function $f(x) = (x^2 + 12)(9 - x^2)$, we need to find critical values, intervals of increase/decrease, local maxima/minima, and intervals of concavity. 2. **Find the derivative $f'(x)$ to locate critical values:** Use the product rule: $f'(x) = (x^2 + 12)'(9 - x^2) + (x^2 + 12)(9 - x^2)'$ Calculate derivatives: $(x^2 + 12)' = 2x$ $(9 - x^2)' = -2x$ So, $$f'(x) = 2x(9 - x^2) + (x^2 + 12)(-2x) = 2x(9 - x^2) - 2x(x^2 + 12)$$ Simplify: $$f'(x) = 2x(9 - x^2 - x^2 - 12) = 2x(9 - 2x^2 - 12) = 2x(-2x^2 - 3) = -4x^3 - 6x$$ 3. **Find critical values by setting $f'(x) = 0$:** $$-4x^3 - 6x = 0$$ Factor out $-2x$: $$-2x(2x^2 + 3) = 0$$ Set each factor to zero: $$-2x = 0 \Rightarrow x = 0$$ $$2x^2 + 3 = 0 \Rightarrow x^2 = -\frac{3}{2}$$ (no real solution) So, the only critical value is $x = 0$. 4. **Determine intervals of increase/decrease using $f'(x)$:** Test points in intervals $(-\infty, 0)$ and $(0, \infty)$: - For $x = -1$, $f'(-1) = -4(-1)^3 - 6(-1) = 4 + 6 = 10 > 0$ (increasing) - For $x = 1$, $f'(1) = -4(1)^3 - 6(1) = -4 - 6 = -10 < 0$ (decreasing) Thus, $f(x)$ is increasing on $(-I, 0)$ and decreasing on $(0, I)$. 5. **Find local maxima and minima using the first derivative test:** At $x=0$, $f'(x)$ changes from positive to negative, so $x=0$ is a local maximum. 6. **Find the second derivative $f''(x)$ to analyze concavity:** $$f''(x) = \frac{d}{dx}(-4x^3 - 6x) = -12x^2 - 6$$ 7. **Determine concavity:** - $f''(x) > 0$ means concave up - $f''(x) < 0$ means concave down Since $-12x^2 - 6$ is always negative (because $-12x^2 \leq 0$ and $-6$ is negative), $f''(x) < 0$ for all $x$. Therefore, $f(x)$ is concave down on $(-I, I)$ and never concave up. **Final answers:** (A) Critical values: $0$ (B) Increasing: $(-I, 0)$ (C) Decreasing: $(0, I)$ (D) Local maxima at $x = 0$ (E) Local minima: $-1000$ (none) (F) Concave up: ${}$ (none) (G) Concave down: $(-I, I)$