The function `f(x) = tan ^-1 x - In(1 + x^2)` is increasing for

Options

A. `x lt (1)/(2)`
B. `x gt (1)/(2)`
C. `x in [-(1)/(2),(1)/(2)]`
D. `x in [-(1)/(2), oo)`

Answer

To determine the intervals where the function \( f(x) = \tan^{-1} x - \ln(1 + x^2) \) is increasing, we need to find its derivative and analyze where this derivative is greater than or equal to zero.

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

The derivative of \( f(x) \) is given by:

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

Using the known derivatives:

- The derivative of \( \tan^{-1} x \) is \( \frac{1}{1 + x^2} \).
- The derivative of \( \ln(1 + x^2) \) is \( \frac{1}{1 + x^2} \cdot 2x = \frac{2x}{1 + x^2} \).

Thus, we have:

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

### Step 2: Simplify the derivative

Combining the terms over a common denominator:

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

### Step 3: Determine where \( f'(x) \geq 0 \)

For the function \( f(x) \) to be increasing, we need:

\[ f'(x) \geq 0 \]

This implies:

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

Since \( 1 + x^2 \) is always positive for all \( x \), we only need to consider the numerator:

\[ 1 - 2x \geq 0 \]

### Step 4: Solve the inequality

Solving the inequality \( 1 - 2x \geq 0 \):

\[ 1 \geq 2x \quad \Rightarrow \quad \frac{1}{2} \geq x \quad \Rightarrow \quad x \leq \frac{1}{2} \]

### Step 5: Conclusion

Thus, the function \( f(x) \) is increasing for:

\[ x \in (-\infty, \frac{1}{2}] \]

### Final Answer:

The function \( f(x) = \tan^{-1} x - \ln(1 + x^2) \) is increasing for \( x \in (-\infty, \frac{1}{2}] \).

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