Range of the function `f (x)= sqrt(x ^(2) + x +1)` is equal to

Options

A. `[0,oo]`
B. `[(sqrt3)/(2), oo)`
C. `((-sqrt3)/(2),(sqrt3)/(2))`
D. `(0,0)`

Answer

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

### Step 1: Analyze the function
We start with the function:
\[ f(x) = \sqrt{x^2 + x + 1} \]
We need to analyze the expression inside the square root, \( x^2 + x + 1 \).

### Step 2: Complete the square
To find the minimum value of \( x^2 + x + 1 \), we can complete the square:
\[ x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + 1 \]
This simplifies to:
\[ \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]

### Step 3: Determine the minimum value
The expression \( \left(x + \frac{1}{2}\right)^2 \) is always non-negative (i.e., it is \( \geq 0 \)). Therefore, the minimum value of \( x^2 + x + 1 \) occurs when \( \left(x + \frac{1}{2}\right)^2 = 0 \), which gives:
\[ x^2 + x + 1 \geq \frac{3}{4} \]

### Step 4: Find the minimum value of \( f(x) \)
Now, substituting the minimum value back into \( f(x) \):
\[ f(x) \geq \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 5: Determine the maximum value
As \( x \) approaches infinity or negative infinity, \( x^2 + x + 1 \) approaches infinity. Therefore, \( f(x) \) also approaches infinity:
\[ f(x) \to \infty \text{ as } x \to \pm \infty \]

### Step 6: Conclude the range
Combining the results from Steps 4 and 5, we find that the range of \( f(x) \) is:
\[ \left[\frac{\sqrt{3}}{2}, \infty\right) \]

### Final Answer
The range of the function \( f(x) = \sqrt{x^2 + x + 1} \) is:
\[ \left[\frac{\sqrt{3}}{2}, \infty\right) \]

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