Range of the function `f(x)= x cos( 1/x), xgt1`

Options

A. `[cos1+sin1 , infty)`
B. `[1, infty)`
C. `[cos1,1)`
D. `[cos1, infty)`

Answer

To find the range of the function \( f(x) = x \cos\left(\frac{1}{x}\right) \) for \( x > 1 \), we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}\left(x \cos\left(\frac{1}{x}\right)\right) \]

Using the product rule, we have:

\[ f'(x) = \cos\left(\frac{1}{x}\right) + x \cdot \frac{d}{dx}\left(\cos\left(\frac{1}{x}\right)\right) \]

Now, we need to differentiate \( \cos\left(\frac{1}{x}\right) \):

\[ \frac{d}{dx}\left(\cos\left(\frac{1}{x}\right)\right) = -\sin\left(\frac{1}{x}\right) \cdot \frac{d}{dx}\left(\frac{1}{x}\right) = -\sin\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = \frac{\sin\left(\frac{1}{x}\right)}{x^2} \]

Thus, we can write:

\[ f'(x) = \cos\left(\frac{1}{x}\right) + x \cdot \frac{\sin\left(\frac{1}{x}\right)}{x^2} = \cos\left(\frac{1}{x}\right) + \frac{\sin\left(\frac{1}{x}\right)}{x} \]

### Step 2: Analyze the Sign of the Derivative
Next, we need to analyze the sign of \( f'(x) \) for \( x > 1 \).

Since \( \cos\left(\frac{1}{x}\right) \) is positive for \( x > 1 \) (as \( \frac{1}{x} < \frac{\pi}{2} \)), and \( \frac{\sin\left(\frac{1}{x}\right)}{x} \) is also positive for \( x > 1 \), we conclude that:

\[ f'(x) > 0 \quad \text{for } x > 1 \]

This means that \( f(x) \) is an increasing function for \( x > 1 \).

### Step 3: Find the Minimum Value
To find the minimum value of \( f(x) \), we evaluate it at the lower bound of the domain:

\[ f(1) = 1 \cdot \cos(1) = \cos(1) \]

### Step 4: Determine the Range
Since \( f(x) \) is increasing for \( x > 1 \) and the minimum value occurs at \( x = 1 \), the range of \( f(x) \) is:

\[ [\cos(1), \infty) \]

### Conclusion
Thus, the range of the function \( f(x) = x \cos\left(\frac{1}{x}\right) \) for \( x > 1 \) is:

\[ \text{Range} = [\cos(1), \infty) \]