To solve this problem, we need to analyze the function:

$f(x) = 2 \sin(x) + \frac{1}{\sqrt{2}} x^2$

The questions involve finding points of inflection and determining the concavity of $f(x)$ on specific intervals.

Steps to Solve the Problem

1. Find the Second Derivative:

   • The first derivative $f'(x)$ will help us locate critical points.
   • The second derivative $f''(x)$ will be used to determine points of inflection, which occur where $f''(x) = 0$ or $f''(x)$ is undefined, and where the concavity changes.

2. Identify Inflection Points:

   • Set $f''(x) = 0$ and solve for $x$ in the interval $[0, 2\pi]$.
   • The values $x = A$ and $x = B$ will be the inflection points within this interval.

3. Determine Concavity on Each Interval:

   • Once $A$ and $B$ are identified, we can test the concavity on the intervals $[0, A]$, $(A, B)$, and $(B, 2\pi]$ by checking the sign of $f''(x)$ within each interval.

I'll proceed with the calculations for each step.Here are the results of our calculations:

1. First Derivative $f'(x)$: $f'(x) = \sqrt{2} x + 2 \cos(x)$

2. Second Derivative $f''(x)$: $f''(x) = -2 \sin(x) + \sqrt{2}$

3. Inflection Points: Solving $f''(x) = 0$ yields two solutions within the interval $[0, 2\pi]$: $A = \frac{\pi}{4} \quad \text{and} \quad B = \frac{3\pi}{4}$

Concavity Analysis

To determine the concavity in each interval $[0, A]$, $(A, B)$, and $(B, 2\pi]$, we will check the sign of $f''(x)$ in these intervals.

Let me perform the concavity test for each interval.The concavity for each interval is as follows:

1. Interval $[0, A] = [0, \frac{\pi}{4}]$:

   • $f''(x) > 0$ (tested at $x = \frac{\pi}{8}$)
   • Therefore, $f(x)$ is concave up on this interval.

2. Interval $(A, B) = \left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$:

   • $f''(x) < 0$ (tested at $x = \frac{\pi}{2}$)
   • Therefore, $f(x)$ is concave down on this interval.

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

   • $f''(x) > 0$ (tested at $x = \pi$)
   • Therefore, $f(x)$ is concave up on this interval.

Summary of Answers

• $A = \frac{\pi}{4}$
• $B = \frac{3\pi}{4}$
• In the interval $[0, A]$, the function $f$ is concave up.
• In the interval $(A, B)$, the function $f$ is concave down.
• In the interval $(B, 2\pi]$, the function $f$ is concave up.

Would you like further details or have any additional questions?

Additional Related Questions

1. How do we determine concavity by examining the sign of the second derivative?
2. What other methods can we use to find inflection points?
3. Can we apply similar techniques to determine concavity on functions with exponential terms?
4. What is the significance of inflection points in real-world applications?
5. How does the behavior of $f(x)$ change around points of inflection?

Tip

When analyzing concavity, choosing test points within each interval helps to determine the behavior of the function on those intervals accurately.