The function f(x) = tan x - x

Options

A. always increases
B. always decreases
C. remains constant
D. sometimes increases and sometimes decreases

Answer

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

### Step 1: Find the derivative of \( f(x) \)

To analyze the behavior of the function, we first need to find its derivative \( f'(x) \).

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

Using the rules of differentiation, we know that:

- The derivative of \( \tan x \) is \( \sec^2 x \).
- The derivative of \( x \) is \( 1 \).

Thus, we have:

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

### Step 2: Analyze the derivative \( f'(x) \)

Next, we need to analyze the expression \( \sec^2 x - 1 \).

Recall that:

\[ \sec^2 x = 1 + \tan^2 x \]

So, we can rewrite the derivative as:

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

### Step 3: Determine the sign of \( f'(x) \)

Since \( \tan^2 x \) is always non-negative (i.e., \( \tan^2 x \geq 0 \) for all \( x \)), we can conclude that:

\[ f'(x) \geq 0 \quad \text{for all } x \]

### Step 4: Conclusion about the function \( f(x) \)

Since the derivative \( f'(x) \) is always non-negative, this implies that the function \( f(x) \) is always increasing.

Thus, the correct option is:

**Option A: The function is always increasing.**

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