Find the range of `f(x)=tan^(-1)sqrt((x^2-2x+2))`

Explanation

To find the range of the function \( f(x) = \tan^{-1}(\sqrt{x^2 - 2x + 2}) \), we will follow these steps:

### Step 1: Simplify the expression inside the square root
The expression \( x^2 - 2x + 2 \) can be rewritten by completing the square:

\[ x^2 - 2x + 2 = (x - 1)^2 + 1 \]

### Step 2: Analyze the square root
Now, we can analyze the square root:

\[ \sqrt{x^2 - 2x + 2} = \sqrt{(x - 1)^2 + 1} \]

Since \( (x - 1)^2 \) is always non-negative, the minimum value of \( (x - 1)^2 + 1 \) is 1 (when \( x = 1 \)). Therefore, we have:

\[ \sqrt{(x - 1)^2 + 1} \geq 1 \]

### Step 3: Determine the range of the square root function
The expression \( \sqrt{(x - 1)^2 + 1} \) can take values from 1 to \( \infty \) as \( x \) varies over all real numbers. Thus, we can write:

\[ \sqrt{x^2 - 2x + 2} \in [1, \infty) \]

### Step 4: Apply the arctangent function
Next, we apply the arctangent function to the range we found:

\[ f(x) = \tan^{-1}(\sqrt{x^2 - 2x + 2}) \]

Since \( \sqrt{x^2 - 2x + 2} \) ranges from 1 to \( \infty \), we find the corresponding values of \( f(x) \):

- When \( \sqrt{x^2 - 2x + 2} = 1 \):
\[ f(x) = \tan^{-1}(1) = \frac{\pi}{4} \]

- As \( \sqrt{x^2 - 2x + 2} \) approaches \( \infty \):
\[ f(x) = \tan^{-1}(\infty) = \frac{\pi}{2} \]

### Step 5: Combine the results
Thus, the range of the function \( f(x) \) is:

\[ \left[ \frac{\pi}{4}, \frac{\pi}{2} \right) \]

### Final Answer
The range of the function \( f(x) = \tan^{-1}(\sqrt{x^2 - 2x + 2}) \) is:

\[ \left[ \frac{\pi}{4}, \frac{\pi}{2} \right) \]

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