1. The problem asks to determine the nature of the critical points at $x=0$ and $x=6.949$ for the function $f$ given its second derivative $$f''(x) = \sin\left(\frac{x^2}{8}\right) - 2\cos x.$$ 2. Recall the second derivative test: - If $f''(x) > 0$ at a critical point, $f$ has a local minimum there. - If $f''(x) < 0$ at a critical point, $f$ has a local maximum there. - If $f''(x) = 0$, the test is inconclusive. 3. 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 $x=0$. 4. Evaluate $f''(x)$ at $x=6.949$: Calculate each term: $$\frac{(6.949)^2}{8} = \frac{48.28}{8} = 6.035.$$ Then $$f''(6.949) = \sin(6.035) - 2\cos(6.949).$$ Using approximate values: $$\sin(6.035) \approx -0.247, \quad \cos(6.949) \approx 0.768.$$ So $$f''(6.949) \approx -0.247 - 2 \times 0.768 = -0.247 - 1.536 = -1.783.$$ Since $f''(6.949) < 0$, $f$ has a local maximum at $x=6.949$. 5. Conclusion: $f$ has a local maximum at both $x=0$ and $x=6.949$. **Final answer:** $f$ has a local maximum at $x=0$ and at $x=6.949$.