If function f `R to R ` is defined by `f (x) = 2x + cos x`, then

Options

A. f has a minimum at `x = pi`
B. has a maximum at x = 0
C. f is a decreasing function
D. f is an increasing function

Answer

To solve the problem step by step, we need to analyze the function \( f(x) = 2x + \cos x \) and determine its behavior in terms of increasing or decreasing.

### Step 1: Differentiate the function
To find out whether the function is increasing or decreasing, we first need to compute the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2x + \cos x) \]

Using the rules of differentiation, we differentiate each term:

- The derivative of \( 2x \) is \( 2 \).
- The derivative of \( \cos x \) is \( -\sin x \).

Thus, we have:

\[ f'(x) = 2 - \sin x \]

### Step 2: Analyze the derivative
Next, we need to analyze the behavior of the derivative \( f'(x) = 2 - \sin x \).

The sine function, \( \sin x \), oscillates between -1 and 1 for all \( x \). Therefore, we can find the minimum and maximum values of \( f'(x) \):

- The maximum value of \( \sin x \) is \( 1 \), which gives us:
\[ f'(x)_{\text{max}} = 2 - 1 = 1 \]
- The minimum value of \( \sin x \) is \( -1 \), which gives us:
\[ f'(x)_{\text{min}} = 2 - (-1) = 3 \]

### Step 3: Determine the sign of the derivative
Since \( f'(x) \) varies between \( 1 \) and \( 3 \), we can conclude that:

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

### Step 4: Conclusion about the function
Since the derivative \( f'(x) \) is always positive, we can conclude that the function \( f(x) \) is always increasing.

### Final Answer
Thus, the function \( f(x) = 2x + \cos x \) is an increasing function.

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