To solve the given problem, we are analyzing the function $f(x) = x + 2\sin x$ over the interval $[0, 2\pi]$. Let’s break this problem into steps:

1. Find the first derivative $f'(x)$:

We differentiate $f(x) = x + 2\sin x$: [ f'(x) = \frac{d}{dx}[x] + \frac{d}{dx}[2\sin x] = 1 + 2\cos x. ]

The first derivative $f'(x)$ helps us determine where the function is increasing or decreasing.

2. Find critical points:

To find critical points, set $f'(x) = 0$: $1 + 2\cos x = 0.$ Simplify: $2\cos x = -1 \quad \implies \quad \cos x = -\frac{1}{2}.$ From the unit circle, $\cos x = -\frac{1}{2}$ at: $x = \frac{2\pi}{3}, \quad x = \frac{4\pi}{3}.$

Thus, the critical points are: $x = \frac{2\pi}{3} \quad \text{and} \quad x = \frac{4\pi}{3}.$

3. Determine where $f(x)$ is increasing or decreasing:

Analyze the sign of $f'(x) = 1 + 2\cos x$ in the intervals determined by the critical points:

• Interval $[0, \frac{2\pi}{3}]$:

  • At $x = 0$, $\cos 0 = 1$, so $f'(0) = 1 + 2(1) = 3 > 0$ (positive).
  • $f(x)$ is increasing on $[0, \frac{2\pi}{3}]$.

• Interval $\left( \frac{2\pi}{3}, \frac{4\pi}{3} \right)$:

  • Here $\cos x < -\frac{1}{2}$, so $1 + 2\cos x < 0$ (negative).
  • $f(x)$ is decreasing on $\left( \frac{2\pi}{3}, \frac{4\pi}{3} \right)$.

• Interval $\left( \frac{4\pi}{3}, 2\pi \right]$:

  • At $x = 2\pi$, $\cos 2\pi = 1$, so $f'(2\pi) = 1 + 2(1) = 3 > 0$ (positive).
  • $f(x)$ is increasing on $\left( \frac{4\pi}{3}, 2\pi \right]$.

4. Find the second derivative $f''(x)$:

The second derivative determines concavity: [ f''(x) = \frac{d}{dx}[f'(x)] = \frac{d}{dx}[1 + 2\cos x] = -2\sin x. ]

5. Find inflection points:

To find where the concavity changes, set $f''(x) = 0$: $-2\sin x = 0 \quad \implies \quad \sin x = 0.$ From the unit circle, $\sin x = 0$ at: $x = 0, \quad x = \pi, \quad x = 2\pi.$

These are candidates for inflection points. Check concavity in intervals:

• Interval $(0, \pi)$:

  • $\sin x > 0$, so $f''(x) = -2\sin x < 0$ (concave down).

• Interval $(\pi, 2\pi)$:

  • $\sin x < 0$, so $f''(x) = -2\sin x > 0$ (concave up).

Thus, the concavity changes at $x = \pi$, making $x = \pi$ an inflection point.

6. Summary of results:

• Increasing: $[0, \frac{2\pi}{3}]$ and $\left[ \frac{4\pi}{3}, 2\pi \right]$.
• Decreasing: $\left( \frac{2\pi}{3}, \frac{4\pi}{3} \right)$.
• Concave down: $(0, \pi)$.
• Concave up: $(\pi, 2\pi)$.
• Inflection point: $x = \pi$.

These conclusions can be confirmed with the graph of $f(x) = x + 2\sin x$.

Would you like a detailed graph explanation or further clarification?

Related Questions:

1. How can critical points help determine local maxima and minima?
2. Why does the second derivative test determine concavity?
3. What happens to $f(x)$ when the first derivative is zero?
4. How can we use trigonometric properties to analyze derivatives?
5. What are inflection points, and why do they matter?

Tip:

Always test the sign of derivatives between critical points to ensure accurate intervals for increasing/decreasing and concavity.