1. The problem asks us to determine the nature of critical points of the function $f$ at $x=0$ and $x=6.949$ using the second derivative test. 2. The second derivative test states: - If $f''(x) > 0$ at a critical point, then $f$ has a local minimum there. - If $f''(x) < 0$ at a critical point, then $f$ has a local maximum there. - If $f''(x) = 0$, the test is inconclusive. 3. Given: $$f''(x) = \sin\left(\frac{x^2}{8}\right) - 2 \cos x$$ 4. Evaluate $f''(x)$ at $x=0$: $$f''(0) = \sin\left(\frac{0^2}{8}\right) - 2 \cos 0 = \sin(0) - 2 \times 1 = 0 - 2 = -2$$ Since $f''(0) = -2 < 0$, $f$ has a local maximum at $x=0$. 5. Evaluate $f''(x)$ at $x=6.949$: Calculate each term: $$\frac{(6.949)^2}{8} = \frac{48.28}{8} = 6.035$$ $$\sin(6.035) \approx -0.247$$ $$\cos(6.949) \approx 0.768$$ So, $$f''(6.949) = -0.247 - 2 \times 0.768 = -0.247 - 1.536 = -1.783$$ Since $f''(6.949) < 0$, $f$ has a local maximum at $x=6.949$. 6. Conclusion: Both critical points correspond to local maxima. Answer: D) $f$ has a local maximum at $x=0$ and at $x=6.949$.