The function f(x)=x+cos x is

Options

A. Always increasing
B. Always decreasing
C. Increasing for certain range of x
D. None of these

Answer

To determine whether the function \( f(x) = x + \cos x \) is increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we first need to find its derivative.

\[ f'(x) = \frac{d}{dx}(x + \cos x) \]

Using the rules of differentiation, we have:

\[ f'(x) = 1 - \sin x \]

### Step 2: Analyze the derivative
Next, we need to determine when the derivative \( f'(x) \) is greater than or equal to 0 (indicating that the function is increasing) or less than 0 (indicating that the function is decreasing).

\[ f'(x) = 1 - \sin x \]

### Step 3: Determine the range of \( \sin x \)
The sine function oscillates between -1 and 1. Therefore, we can analyze the expression \( 1 - \sin x \):

- The maximum value of \( \sin x \) is 1, which gives:
\[ f'(x) = 1 - 1 = 0 \]

- The minimum value of \( \sin x \) is -1, which gives:
\[ f'(x) = 1 - (-1) = 1 + 1 = 2 \]

### Step 4: Conclude the behavior of the function
Since \( \sin x \) ranges from -1 to 1, the derivative \( f'(x) \) will always be:

\[ f'(x) \geq 0 \quad \text{for all } x \]

This indicates that \( f'(x) \) is non-negative for all \( x \), meaning that the function \( f(x) \) is always increasing.

### Final Conclusion
Thus, the function \( f(x) = x + \cos x \) is an increasing function for all \( x \in \mathbb{R} \).

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