The function `f(x)=tanx-x`

Options

A. always increases
B. always decreases
C. never decreases
D. sometimes increases and sometimes decreases

Explanation

To determine whether the function \( f(x) = \tan x - x \) is increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
We start by calculating the derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(\tan x) - \frac{d}{dx}(x) \]

Using the derivative of \( \tan x \) which is \( \sec^2 x \) and the derivative of \( x \) which is \( 1 \), we have:

\[ f'(x) = \sec^2 x - 1 \]

### Step 2: Simplify the derivative
Next, we can simplify the expression for the derivative:

\[ f'(x) = \sec^2 x - 1 = \tan^2 x \]

This is because \( \sec^2 x = 1 + \tan^2 x \).

### Step 3: Analyze the sign of the derivative
Now, we need to analyze the sign of \( f'(x) \):

- The function \( \tan^2 x \) is always non-negative for all \( x \) where it is defined.
- Therefore, \( f'(x) \geq 0 \) for all \( x \) in the domain of \( f(x) \).

### Step 4: Conclusion about the function
Since \( f'(x) \geq 0 \), this implies that the function \( f(x) \) is always increasing in its domain.

### Final Answer
The function \( f(x) = \tan x - x \) is always increasing.

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