2.9 Q-3

Question Statement

Find the maximum and minimum values of the function $f(x) = \sin x + \cos x$ in the interval $[0, 2\pi]$.

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Background and Explanation

To solve this problem, we need to determine the critical points and evaluate the function at these points and the endpoints of the interval. Key concepts include:

• Critical Points: These occur where the derivative of the function is zero or undefined.
• Trigonometric Properties: Knowing that $\sin x + \cos x$ can be rewritten using trigonometric identities helps simplify computations.

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Solution

We begin by analyzing the function $f(x) = \sin x + \cos x$.

Step 1: Simplify the Function

Using the identity $\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$, the function’s behavior depends on $\sin\left(x + \frac{\pi}{4}\right)$, which oscillates between $-1$ and $1$. However, we solve this using derivatives to verify.

Step 2: Find the Derivative

The derivative of $f(x)$ is:

$$f'(x) = \cos x - \sin x$$

To find critical points, set $f'(x) = 0$:

$$\cos x - \sin x = 0 \quad \Rightarrow \quad \cos x = \sin x$$

This occurs when $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$ in $[0, 2\pi]$.

Step 3: Analyze the Function at Critical Points

At $x = \frac{\pi}{4}$:

The value of $f(x)$ is:

$$f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$$

At $x = \frac{5\pi}{4}$:

The value of $f(x)$ is:

$$f\left(\frac{5\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}$$

Step 4: Evaluate at the Endpoints

At $x = 0$:

$$f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$$

At $x = 2\pi$:

$$f(2\pi) = \sin(2\pi) + \cos(2\pi) = 0 + 1 = 1$$

Step 5: Determine Maximum and Minimum Values

From the above calculations:

• The maximum value is $\sqrt{2}$, occurring at $x = \frac{\pi}{4}$.
• The minimum value is $-\sqrt{2}$, occurring at $x = \frac{5\pi}{4}$.

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Key Formulas or Methods Used

1. Derivative of trigonometric functions:

$$\frac{d}{dx} (\sin x) = \cos x, \quad \frac{d}{dx} (\cos x) = -\sin x$$

2. Critical points occur where $f'(x) = 0$ or is undefined.
3. Simplification using trigonometric identities:

$$\sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$$

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Summary of Steps

1. Compute the derivative $f'(x) = \cos x - \sin x$.
2. Solve $f'(x) = 0$ to find critical points: $x = \frac{\pi}{4}, \frac{5\pi}{4}$.
3. Evaluate $f(x)$ at critical points and endpoints ($x = 0, 2\pi$).
4. Compare values to determine:
   • Maximum value: $\sqrt{2}$ at $x = \frac{\pi}{4}$.
   • Minimum value: $-\sqrt{2}$ at $x = \frac{5\pi}{4}$.