consider the function `f:R rarr R,f(x)=(x^(2)-6x+4)/(x^(2)+2x+4)`
f(x) is

Options

A. unbounded function
B. one one function
C. onto function
D. none of these

Explanation

To analyze the function \( f(x) = \frac{x^2 - 6x + 4}{x^2 + 2x + 4} \) and determine its properties, we will follow these steps:

### Step 1: Simplify the function
We can rewrite the function in a more manageable form. We notice that we can express the numerator in terms of the denominator:

\[ f(x) = \frac{x^2 - 6x + 4}{x^2 + 2x + 4} \]

We can rewrite the numerator as:

\[ x^2 - 6x + 4 = (x^2 + 2x + 4) - 8x \]

Thus, we have:

\[ f(x) = \frac{(x^2 + 2x + 4) - 8x}{x^2 + 2x + 4} = 1 - \frac{8x}{x^2 + 2x + 4} \]

### Step 2: Differentiate the function
Next, we need to find the derivative \( f'(x) \) using the quotient rule. The function can be expressed as:

\[ f(x) = 1 - \frac{8x}{x^2 + 2x + 4} \]

Let \( g(x) = 8x \) and \( h(x) = x^2 + 2x + 4 \). The derivative using the quotient rule is:

\[ f'(x) = -\frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

Calculating \( g'(x) = 8 \) and \( h'(x) = 2x + 2 \):

\[ f'(x) = -\frac{8(x^2 + 2x + 4) - 8x(2x + 2)}{(x^2 + 2x + 4)^2} \]

### Step 3: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):

\[ 8(x^2 + 2x + 4) - 8x(2x + 2) = 0 \]

This simplifies to:

\[ 8(x^2 + 2x + 4 - 2x^2 - 2x) = 0 \]

\[ 8(-x^2 + 4) = 0 \implies -x^2 + 4 = 0 \implies x^2 = 4 \]

Thus, \( x = -2 \) and \( x = 2 \).

### Step 4: Evaluate the function at critical points
Now we evaluate \( f(x) \) at these critical points:

1. For \( x = -2 \):

\[ f(-2) = \frac{(-2)^2 - 6(-2) + 4}{(-2)^2 + 2(-2) + 4} = \frac{4 + 12 + 4}{4 - 4 + 4} = \frac{20}{4} = 5 \]

2. For \( x = 2 \):

\[ f(2) = \frac{(2)^2 - 6(2) + 4}{(2)^2 + 2(2) + 4} = \frac{4 - 12 + 4}{4 + 4 + 4} = \frac{-4}{12} = -\frac{1}{3} \]

### Step 5: Determine the range of the function
From the evaluations, we find that:

- \( f(-2) = 5 \)
- \( f(2) = -\frac{1}{3} \)

Thus, the range of the function is:

\[ \text{Range of } f(x) = \left[-\frac{1}{3}, 5\right] \]

### Step 6: Conclusion about the function
Since the range is bounded, we conclude that:

- The function is **bounded**.
- It is **not one-one** (since it takes multiple values in the range).
- It is **not onto** (since it does not cover all real numbers).

Thus, the answer is **none of these**.

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