1. **Problem statement:** We are given the second derivative of a function $f$ as $$f''(x) = x^2 \cos\left(\frac{x^2 + 2x}{6}\right)$$ and asked to find the values of $x$ in the interval $(-4,3)$ where the graph of $f$ has points of inflection. 2. **Recall:** Points of inflection occur where the second derivative changes sign, which typically happens where $$f''(x) = 0$$ or where $f''(x)$ is undefined (not the case here). 3. **Set the second derivative equal to zero:** $$ x^2 \cos\left(\frac{x^2 + 2x}{6}\right) = 0 $$ This product equals zero if either factor is zero: - $x^2 = 0$ - $\cos\left(\frac{x^2 + 2x}{6}\right) = 0$ 4. **Solve $x^2 = 0$:** $$ x = 0 $$ 5. **Solve $\cos\left(\frac{x^2 + 2x}{6}\right) = 0$:** Recall that $\cos \theta = 0$ at $$\theta = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$ Set: $$ \frac{x^2 + 2x}{6} = \frac{\pi}{2} + k\pi $$ Multiply both sides by 6: $$ x^2 + 2x = 3\pi + 6k\pi $$ Rewrite as: $$ x^2 + 2x - 3\pi - 6k\pi = 0 $$ 6. **Solve quadratic for $x$:** $$ x = \frac{-2 \pm \sqrt{4 + 4(3\pi + 6k\pi)}}{2} = -1 \pm \sqrt{1 + 3\pi + 6k\pi} $$ 7. **Find integer values of $k$ such that $x$ lies in $(-4,3)$:** - For $k=0$: $$ x = -1 \pm \sqrt{1 + 3\pi} \approx -1 \pm 3.23 $$ So, $$ x_1 \approx -1 - 3.23 = -4.23 \quad (\text{outside } (-4,3)) $$ $$ x_2 \approx -1 + 3.23 = 2.23 \quad (\text{inside } (-4,3)) $$ - For $k=-1$: $$ x = -1 \pm \sqrt{1 + 3\pi - 6\pi} = -1 \pm \sqrt{1 - 3\pi} $$ Since $3\pi \approx 9.42$, $1 - 9.42 < 0$, no real solutions. - For $k=1$: $$ x = -1 \pm \sqrt{1 + 3\pi + 6\pi} = -1 \pm \sqrt{1 + 9\pi} $$ Since $9\pi \approx 28.27$, $$ x = -1 \pm 5.44 $$ So, $$ x_1 = -1 - 5.44 = -6.44 \quad (\text{outside } (-4,3)) $$ $$ x_2 = -1 + 5.44 = 4.44 \quad (\text{outside } (-4,3)) $$ No other $k$ values will produce $x$ in $(-4,3)$. 8. **Summary of solutions in $(-4,3)$:** $$ x = 0 \quad \text{and} \quad x \approx 2.23 $$ 9. **Conclusion:** The points of inflection occur at $x=0$ and $x \approx 2.229$ (rounded). **Answer choice:** B) 0 and 2.229