Problem Analysis #20

To find the maximum $M$ and minimum $m$ of the function $f(\theta) = \sin\theta - \theta + \frac{\theta^3}{6}$ on the interval $-1 \leq \theta \leq 1$, we analyze the function's behavior through its derivative and critical points.

Step 1: Compute the derivative.
The derivative of $f(\theta)$ is:

Step 2: Find critical points.
Set $f'(\theta) = 0$:

This equation is transcendental and doesn't have a simple closed-form solution. However, we can analyze the behavior of $\cos\theta$ and $1 - \frac{\theta^2}{2}$ on $[-1, 1]$.

Step 3: Evaluate endpoints and critical points.

• At $\theta = -1$: $$f(-1) = \sin(-1) - (-1) + \frac{(-1)^3}{6} = -\sin1 + 1 - \frac{1}{6} = 1 - \sin1 - \frac{1}{6}.$$
• At $\theta = 1$: $$f(1) = \sin1 - 1 + \frac{1^3}{6} = \sin1 - 1 + \frac{1}{6} = \sin1 - \frac{5}{6}.$$

Step 4: Determine the nature of critical points.
Since $\cos\theta$ is concave down on $[-1, 1]$ and $1 - \frac{\theta^2}{2}$ is concave up, they intersect exactly once in $(-1, 1)$. This implies a single critical point where $f'(\theta) = 0$, which is a local extremum.

Step 5: Analyze the function's behavior.

• The function $f(\theta)$ is continuous and differentiable on $[-1, 1]$.
• The derivative $f'(\theta)$ changes sign at the critical point, indicating a local maximum or minimum.
• The endpoints and the critical point are the only candidates for $M$ and $m$.

Step 6: Compute $M + m$.
Since the critical point is the only extremum, and the function is continuous, the sum $M + m$ is the sum of the function values at the endpoints and the critical point. However, due to the transcendental nature of the critical point, we approximate numerically or use symmetry properties.

Given the symmetry and the transcendental nature, the sum $M + m$ simplifies to a specific integer value based on the function's properties and the interval's endpoints.

Final Answer: