If `f(x)=(tan^(-1)x)^(2) +(2)/(sqrt(x^(2)+1))` then f(x) is increasing in

Options

A. `(0, oo)`
B. `(-oo, 0)`
C. `(-oo, -5)`
D. None of these

Answer

To determine where the function \( f(x) = (\tan^{-1} x)^2 + \frac{2}{\sqrt{x^2 + 1}} \) is increasing, we need to find the derivative \( f'(x) \) and analyze its sign.

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

1. The first term is \( (\tan^{-1} x)^2 \). Using the chain rule:
\[ \frac{d}{dx}[(\tan^{-1} x)^2] = 2 \tan^{-1} x \cdot \frac{d}{dx}[\tan^{-1} x] = 2 \tan^{-1} x \cdot \frac{1}{1 + x^2} \]

2. The second term is \( \frac{2}{\sqrt{x^2 + 1}} \). We can rewrite it as \( 2 (x^2 + 1)^{-1/2} \) and differentiate using the chain rule:
\[ \frac{d}{dx}\left[2 (x^2 + 1)^{-1/2}\right] = 2 \cdot -\frac{1}{2} (x^2 + 1)^{-3/2} \cdot \frac{d}{dx}[x^2 + 1] = -\frac{2x}{(x^2 + 1)^{3/2}} \]

Combining these results, we have:
\[ f'(x) = 2 \tan^{-1} x \cdot \frac{1}{1 + x^2} - \frac{2x}{(x^2 + 1)^{3/2}} \]

### Step 2: Set the derivative greater than or equal to zero
To find where \( f(x) \) is increasing, we set \( f'(x) \geq 0 \):
\[ 2 \tan^{-1} x \cdot \frac{1}{1 + x^2} - \frac{2x}{(x^2 + 1)^{3/2}} \geq 0 \]

### Step 3: Simplify the inequality
We can factor out \( 2 \) from the inequality:
\[ \tan^{-1} x \cdot \frac{1}{1 + x^2} \geq \frac{x}{(x^2 + 1)^{3/2}} \]

### Step 4: Analyze the inequality
To analyze the inequality, we can rewrite it as:
\[ \tan^{-1} x \geq \frac{x(1 + x^2)}{(x^2 + 1)^{3/2}} \]
This requires further analysis or numerical methods to find the intervals where this holds true.

### Step 5: Test values
We can test specific values of \( x \) to determine where the inequality holds:

1. For \( x = 0 \):
\[ \tan^{-1}(0) = 0 \quad \text{and} \quad \frac{0(1 + 0^2)}{(0^2 + 1)^{3/2}} = 0 \]
So, \( f'(0) = 0 \).

2. For \( x = 1 \):
\[ \tan^{-1}(1) = \frac{\pi}{4} \quad \text{and} \quad \frac{1(1 + 1^2)}{(1^2 + 1)^{3/2}} = \frac{1 \cdot 2}{(2)^{3/2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} \approx 0.707 \]
Since \( \frac{\pi}{4} \approx 0.785 > 0.707 \), \( f'(1) > 0 \).

3. For larger values of \( x \), we can observe that \( \tan^{-1} x \) approaches \( \frac{\pi}{2} \) while the right side approaches \( 0 \), indicating \( f'(x) > 0 \).

### Conclusion
From our analysis, we conclude that \( f(x) \) is increasing for \( x \geq 0 \).