The function `f(x)=cos x-sinx` has maximum or minimum value at `x=…..`

Options

A. `(pi)/(4)`
B. `(3pi)/(4)`
C. `(pi)/(2)`
D. `(pi)/(3).`

Answer

To find the maximum or minimum value of the function \( f(x) = \cos x - \sin x \), we will follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of the function \( f(x) \).

\[ f'(x) = -\sin x - \cos x \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero.

\[ -\sin x - \cos x = 0 \]

This simplifies to:

\[ \sin x + \cos x = 0 \]

### Step 3: Solve for \( x \)
We can rearrange the equation:

\[ \sin x = -\cos x \]

Dividing both sides by \( \cos x \) (assuming \( \cos x \neq 0 \)) gives:

\[ \tan x = -1 \]

The general solution for \( \tan x = -1 \) is:

\[ x = \frac{3\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Step 4: Determine the nature of the critical points
To determine whether these critical points are maxima or minima, we can use the second derivative test.

First, we find the second derivative:

\[ f''(x) = -\cos x + \sin x \]

Now, we evaluate the second derivative at \( x = \frac{3\pi}{4} \):

\[ f''\left(\frac{3\pi}{4}\right) = -\cos\left(\frac{3\pi}{4}\right) + \sin\left(\frac{3\pi}{4}\right) \]

Calculating the values:

\[ \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \quad \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

Thus,

\[ f''\left(\frac{3\pi}{4}\right) = -\left(-\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} > 0 \]

Since \( f''\left(\frac{3\pi}{4}\right) > 0 \), this indicates that \( x = \frac{3\pi}{4} \) is a local minimum.

### Step 5: Conclusion
Thus, the function \( f(x) = \cos x - \sin x \) has a minimum value at:

\[ x = \frac{3\pi}{4} \]