To solve this problem, let's go through each part of the analysis step by step.

Given Function and Interval

We have: $f(x) = \sin\left(x + \frac{\pi}{4}\right)$ over the interval $0 \leq x \leq 2\pi$.

Step 1: Find Critical Points

To find the critical points, we need to take the derivative of $f(x)$ and set it equal to zero.

1. Derivative: $f'(x) = \cos\left(x + \frac{\pi}{4}\right)$
2. Set the Derivative to Zero: $\cos\left(x + \frac{\pi}{4}\right) = 0$ Solving for $x$, we find: $x + \frac{\pi}{4} = \frac{\pi}{2} + n\pi \quad \text{for integer } n$ This gives: $x = \frac{\pi}{4} + n\pi$ For the interval $0 \leq x \leq 2\pi$, the valid solutions for $x$ are: $x = \frac{\pi}{4}, \frac{5\pi}{4}$

Step 2: Evaluate $f(x)$ at Critical Points and Endpoints

We evaluate $f(x)$ at the critical points $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$, and also at the endpoints $x = 0$ and $x = 2\pi$:

1. At $x = 0$: $f(0) = \sin\left(0 + \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
2. At $x = 2\pi$: $f(2\pi) = \sin\left(2\pi + \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
3. At $x = \frac{\pi}{4}$: $f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{2}\right) = 1$
4. At $x = \frac{5\pi}{4}$: $f\left(\frac{5\pi}{4}\right) = \sin\left(\frac{3\pi}{2}\right) = -1$

Step 3: Determine Absolute and Local Maximum and Minimum

• The absolute maximum of $f(x)$ on this interval is $1$, occurring at $x = \frac{\pi}{4}$.
• The absolute minimum of $f(x)$ is $-1$, occurring at $x = \frac{5\pi}{4}$.
• Since $f(x)$ reaches a global maximum and minimum at these points, there are no additional local maxima or minima in this interval.

Answers to Parts

Part 1: Absolute Maximum

The absolute maximum value is $1$ at $x = \frac{\pi}{4}$. So, the answer is:

• A. The absolute maximum value is $1$ at $x = \frac{\pi}{4}$.

Part 2: Local Maxima which are not Absolute

There are no local maxima that are not absolute because the only maximum value within the interval is the absolute maximum.

So, the answer is:

• C. There are no local maxima which are not absolute.

Would you like more details on any part of this analysis?

Related Questions

1. What are the absolute minimum values of this function in the given interval?
2. How can we identify non-stationary critical points in trigonometric functions?
3. How does changing the interval affect the absolute maximum and minimum of periodic functions?
4. What would the critical points be if the interval was extended to $0 \leq x \leq 4\pi$?
5. How does the phase shift $\frac{\pi}{4}$ affect the critical points of $f(x) = \sin(x)$?

Tip

Remember, for periodic functions like sine and cosine, critical points repeat periodically.