If `f:[3,oo)toR` is a function defined by `f(x)=x^(2)-2x+6`, then the range of f is

Options

A. R
B. `[5,oo)`
C. `[6,oo)`
D. `[9,oo)`

Answer

To find the range of the function \( f(x) = x^2 - 2x + 6 \) defined on the interval \( [3, \infty) \), we can follow these steps:

### Step 1: Rewrite the function in a more manageable form
We start with the function:
\[ f(x) = x^2 - 2x + 6 \]
We can complete the square for the quadratic expression.

### Step 2: Complete the square
To complete the square, we take the first two terms:
\[ x^2 - 2x \]
We can rewrite this as:
\[ (x - 1)^2 - 1 \]
Now, substituting back into the function:
\[ f(x) = (x - 1)^2 - 1 + 6 = (x - 1)^2 + 5 \]

### Step 3: Analyze the function
The expression \( (x - 1)^2 \) is always non-negative since it is a square term. The minimum value occurs when \( x = 1 \), giving:
\[ (x - 1)^2 \geq 0 \]
Thus,
\[ f(x) \geq 5 \]

### Step 4: Determine the minimum value of \( f(x) \) within the domain
Since the function is defined for \( x \geq 3 \), we evaluate \( f(x) \) at the lower bound of the domain:
\[ f(3) = (3 - 1)^2 + 5 = 2^2 + 5 = 4 + 5 = 9 \]
As \( x \) increases beyond 3, \( f(x) \) will only increase since it is a quadratic function that opens upwards.

### Step 5: Conclude the range
Since the minimum value of \( f(x) \) when \( x = 3 \) is 9 and the function increases without bound as \( x \) approaches infinity, the range of the function \( f \) is:
\[ [9, \infty) \]

### Final Answer
The range of the function \( f \) is \( [9, \infty) \).

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