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

Options

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

Answer

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

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( f(x) \).

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

Using the derivative of the inverse tangent function, we know:

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

And the derivative of \( x \) is simply \( 1 \). Therefore, we have:

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

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

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

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

\[ f'(x) > 0 \]

Since both the numerator \( 2 + x^2 \) and the denominator \( 1 + x^2 \) are always positive for all real \( x \) (as \( x^2 \geq 0 \)), we conclude that:

\[ f'(x) > 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 4: Conclusion
Since \( f'(x) > 0 \) for all \( x \), the function \( f(x) \) is increasing on the interval:

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

### Summary
The function \( f(x) = \tan^{-1}(x) + x \) increases in the interval \( (-\infty, \infty) \).

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