The domain of the function f(x)=max{sin x, cos x} is `(-infty, infty)`. The range of f(x) is

Options

A. `[-1/sqrt(2),1]`
B. `[-1/sqrt(2),1/sqrt(2)]`
C. [0,1]
D. [-1,1]

Answer

To find the range of the function \( f(x) = \max(\sin x, \cos x) \), we can follow these steps:

### Step 1: Understand the Functions
The functions involved are \( \sin x \) and \( \cos x \). Both functions oscillate between -1 and 1 for all \( x \).

### Step 2: Analyze the Graphs
To determine the range of \( f(x) \), we can visualize the graphs of \( \sin x \) and \( \cos x \):
- The sine function starts at 0 when \( x = 0 \) and oscillates between -1 and 1.
- The cosine function starts at 1 when \( x = 0 \) and also oscillates between -1 and 1.

### Step 3: Identify Points of Intersection
The points where \( \sin x \) and \( \cos x \) intersect are important because at these points, \( f(x) \) will switch between the two functions. The points of intersection occur when:
\[ \sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
At these points, both functions equal \( \frac{\sqrt{2}}{2} \).

### Step 4: Determine the Maximum Values
- The maximum value of \( \sin x \) is 1, which occurs at \( x = \frac{\pi}{2} + 2n\pi \).
- The maximum value of \( \cos x \) is also 1, which occurs at \( x = 2n\pi \).
- The minimum value of both functions is -1, but since we are taking the maximum, we need to find the minimum point of interest.

### Step 5: Find the Range of \( f(x) \)
The function \( f(x) = \max(\sin x, \cos x) \) will take values from the minimum intersection point to the maximum value:
- The minimum value of \( f(x) \) occurs at the points where both sine and cosine are equal, which is \( \frac{\sqrt{2}}{2} \) (approximately 0.707).
- The maximum value is 1.

Thus, the range of \( f(x) \) is:
\[ \left[\frac{\sqrt{2}}{2}, 1\right] \]

### Final Answer
The range of the function \( f(x) = \max(\sin x, \cos x) \) is:
\[ \left[\frac{\sqrt{2}}{2}, 1\right] \]