The function f (x) =tan x-x

Options

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

Answer

To analyze the function \( f(x) = \tan x - x \) and determine whether it is always increasing, always decreasing, never decreasing, or sometimes increasing and sometimes decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
To determine the behavior of the function, we first need to find its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(\tan x - x) = \sec^2 x - 1 \]

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

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

### 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 since it is a square of the tangent function.
- Therefore, \( f'(x) \geq 0 \) for all \( x \) where \( \tan x \) is defined.

### Step 4: Determine the intervals of increase and decrease
Since \( f'(x) \geq 0 \), this implies that the function \( f(x) \) is non-decreasing in the intervals where it is defined. However, we must also consider the points where \( \tan x \) is undefined (i.e., \( x = \frac{\pi}{2} + n\pi \) for \( n \in \mathbb{Z} \)).

### Step 5: Conclusion
From our analysis, we conclude that:
- The function \( f(x) = \tan x - x \) is always increasing in the intervals where it is defined because \( f'(x) \geq 0 \).

Thus, the correct option is that the function is **always increasing**.

---