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 \( f(x) = \tan^{-1}(x) - x \) decreases, we will follow these steps:

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

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

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

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

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

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

### Step 3: Analyze the sign of the derivative
To find where the function is decreasing, we need to determine where \( f'(x) < 0 \):

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

The numerator \( -x^2 \) is always less than or equal to 0 for all real \( x \) (it is 0 when \( x = 0 \) and negative otherwise). The denominator \( 1 + x^2 \) is always positive for all real \( x \). Therefore, the entire fraction \( \frac{-x^2}{1 + x^2} \) is less than 0 for all \( x \neq 0 \).

### Step 4: Conclusion about the interval of decrease
Since \( f'(x) < 0 \) for all \( x \neq 0 \), we conclude that the function \( f(x) \) is decreasing for all \( x \in (-\infty, \infty) \).

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

---