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

Options

A. `(0, infty)`
B. [1,10]
C. [3,5]
D. [2,5]

Answer

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

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

Given:
\[ f(x) = \left( \tan^{-1} x \right)^2 + \frac{2}{\sqrt{x^2 + 1}} \]

Using the chain rule and quotient rule, we differentiate \( f(x) \).

1. The derivative of \( \left( \tan^{-1} x \right)^2 \) is:
\[ 2 \tan^{-1} x \cdot \frac{d}{dx}(\tan^{-1} x) = 2 \tan^{-1} x \cdot \frac{1}{1 + x^2} \]

2. The derivative of \( \frac{2}{\sqrt{x^2 + 1}} \) using the quotient rule:
\[ \frac{d}{dx} \left( \frac{2}{\sqrt{x^2 + 1}} \right) = 2 \cdot \frac{d}{dx} \left( (x^2 + 1)^{-1/2} \right) = 2 \cdot \left( -\frac{1}{2}(x^2 + 1)^{-3/2} \cdot 2x \right) = -\frac{2x}{(x^2 + 1)^{3/2}} \]

Combining these, 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 \( f'(x) \) greater than or equal to zero

We want to find where:
\[ f'(x) \geq 0 \]
This leads to:
\[ 2 \tan^{-1} x \cdot \frac{1}{1 + x^2} - \frac{2x}{(x^2 + 1)^{3/2}} \geq 0 \]

### Step 3: Factor out common terms

We can factor out \( \frac{2}{1 + x^2} \):
\[ \frac{2}{1 + x^2} \left( \tan^{-1} x - \frac{x}{\sqrt{x^2 + 1}} \right) \geq 0 \]

Since \( \frac{2}{1 + x^2} > 0 \) for all \( x \), we focus on:
\[ \tan^{-1} x - \frac{x}{\sqrt{x^2 + 1}} \geq 0 \]

### Step 4: Analyze the inequality

We need to show that:
\[ \tan^{-1} x \geq \frac{x}{\sqrt{x^2 + 1}} \]

### Step 5: Check the behavior of both sides

1. As \( x \to 0 \):
\[ \tan^{-1}(0) = 0 \quad \text{and} \quad \frac{0}{\sqrt{0^2 + 1}} = 0 \]
Hence, both sides are equal.

2. As \( x \to \infty \):
\[ \tan^{-1}(x) \to \frac{\pi}{2} \quad \text{and} \quad \frac{x}{\sqrt{x^2 + 1}} \to 1 \]
Thus, \( \tan^{-1}(x) > \frac{x}{\sqrt{x^2 + 1}} \).

### Step 6: Conclusion

Since \( \tan^{-1} x \) is always greater than or equal to \( \frac{x}{\sqrt{x^2 + 1}} \) for \( x \geq 0 \), we conclude that \( f'(x) \geq 0 \) for \( x \in [0, \infty) \).

Thus, the function \( f(x) \) is increasing in the interval:
\[ \text{Answer: } (0, \infty) \]