Question

For the function $f(x) = \sin x + \cos x$ , the point of relative maxima on $[0, 2\pi]$ is given by

• A. $(\pi/4, \sqrt{2})$
• B. $(-\pi/4, -\sqrt{2})$
• C. $(-\pi/4, \sqrt{2})$
• D. $(\pi/4, -\sqrt{2})$

Hint:

Any function of the form $a \sin x + b \cos x$ has a maximum value of $\sqrt{a^2 + b^2}$. For $\sin x + \cos x$, the maximum value is $\sqrt{1^2 + 1^2} = \sqrt{2}$, which occurs at $x = \pi/4$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the relative extrema of a function, we locate its critical points by setting the first derivative to zero and then apply the second derivative test.

Step 2: Detailed Explanation:
Given the function: \[ f(x) = \sin x + \cos x \]
First, find the first derivative $f'(x)$ and set it to zero to find the critical points: \[ f'(x) = \cos x - \sin x = 0 \implies \cos x = \sin x \implies \tan x = 1 \]
Within the interval $[0, 2\pi]$, the solutions for $\tan x = 1$ are: \[ x = \frac{\pi}{4} \text{ and } x = \frac{5\pi}{4} \]
Next, compute the second derivative $f''(x)$ to determine the nature of these critical points: \[ f''(x) = -\sin x - \cos x \]
Evaluate the second derivative at $x = \frac{\pi}{4}$: \[ f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2} < 0 \]
Since $f''(x) < 0$ at $x = \frac{\pi}{4}$, the function has a relative maximum at this point.
Calculate the maximum functional value: \[ f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \]
Therefore, the point of relative maximum is $\left(\frac{\pi}{4}, \sqrt{2}\right)$.

Step 3: Final Answer:
The point of relative maxima is $(\pi/4, \sqrt{2})$.