The function `f(x) = tan ^(-1)x -x` decreases in the interval

Options

A. `(1, infty)`
B. `(1,- infty)`
C. `(- infty, infty)`
D. `(0, infty)`

Answer

To determine the interval in which the function \( f(x) = \tan^{-1}(x) - x \) is decreasing, we will follow these steps:

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

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

Using the derivative of \( \tan^{-1}(x) \), which is \( \frac{1}{1+x^2} \), we get:

\[ f'(x) = \frac{1}{1+x^2} - 1 \]

### Step 2: Simplify the derivative
Next, we simplify the expression for \( f'(x) \):

\[ f'(x) = \frac{1 - (1+x^2)}{1+x^2} = \frac{1 - 1 - x^2}{1+x^2} = \frac{-x^2}{1+x^2} \]

### Step 3: Determine when the derivative is less than zero
To find the intervals where the function is decreasing, we need to set the derivative less than zero:

\[ f'(x) < 0 \]

This gives us:

\[ \frac{-x^2}{1+x^2} < 0 \]

### Step 4: Analyze the inequality
The numerator \( -x^2 \) is negative for all \( x \neq 0 \), and the denominator \( 1+x^2 \) is always positive for all real \( x \). Therefore, \( f'(x) < 0 \) for all \( x \neq 0 \).

### Step 5: Conclusion
Since \( f'(x) < 0 \) for all \( x \) except at \( x = 0 \), we conclude that the function \( f(x) \) is decreasing in the interval:

\[ (-\infty, \infty) \]

### Final Answer
The function \( f(x) = \tan^{-1}(x) - x \) decreases in the interval \( (-\infty, \infty) \).

---