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

Options

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

Answer

To determine the interval in which the function \( f(x) = \cot^{-1}(x) + x \) is increasing, we need to find the derivative of the function and analyze its sign.

### Step 1: Find the derivative of the function
The function is given as:
\[ f(x) = \cot^{-1}(x) + x \]
To find the derivative \( f'(x) \), we differentiate each term:
\[ f'(x) = \frac{d}{dx}(\cot^{-1}(x)) + \frac{d}{dx}(x) \]
The derivative of \( \cot^{-1}(x) \) is:
\[ \frac{d}{dx}(\cot^{-1}(x)) = -\frac{1}{1 + x^2} \]
And the derivative of \( x \) is:
\[ \frac{d}{dx}(x) = 1 \]
Thus, we have:
\[ f'(x) = -\frac{1}{1 + x^2} + 1 \]

### Step 2: Simplify the derivative
Now, we simplify \( f'(x) \):
\[ f'(x) = 1 - \frac{1}{1 + x^2} \]
To combine these terms, we find a common denominator:
\[ f'(x) = \frac{(1 + x^2) - 1}{1 + x^2} = \frac{x^2}{1 + x^2} \]

### Step 3: Analyze the sign of the derivative
Next, we analyze the expression \( f'(x) = \frac{x^2}{1 + x^2} \):
- The numerator \( x^2 \) is always non-negative (i.e., \( x^2 \geq 0 \)).
- The denominator \( 1 + x^2 \) is always positive (i.e., \( 1 + x^2 > 0 \) for all real \( x \)).

Thus, \( f'(x) \) is non-negative for all \( x \):
\[ f'(x) \geq 0 \quad \text{for all } x \in \mathbb{R} \]

### Step 4: Determine the intervals of increase
Since \( f'(x) \) is greater than or equal to zero for all \( x \), the function \( f(x) \) is increasing in the entire real line:
\[ \text{The function } f(x) \text{ is increasing in the interval } (-\infty, \infty). \]

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

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