If x is real, then `(x^(2)-2x+4)/(x^(2)+2x+4)` takes values in the interval

Options

A. `[1/3,3]`
B. (1/3,3)
C. (3,3)
D. (-1/3,3)

Answer

To solve the problem of determining the interval in which the function \( f(x) = \frac{x^2 - 2x + 4}{x^2 + 2x + 4} \) takes values when \( x \) is real, we can follow these steps:

### Step 1: Set up the equation
Let \( y = f(x) = \frac{x^2 - 2x + 4}{x^2 + 2x + 4} \).

### Step 2: Multiply both sides by the denominator
Multiply both sides by \( x^2 + 2x + 4 \) (which is always positive for real \( x \)):
\[ y(x^2 + 2x + 4) = x^2 - 2x + 4 \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ yx^2 + 2yx + 4y - x^2 + 2x - 4 = 0 \]
Combine like terms:
\[ (1 - y)x^2 + (2y + 2)x + (4y - 4) = 0 \]

### Step 4: Identify coefficients
The coefficients are:
- \( a = 1 - y \)
- \( b = 2y + 2 \)
- \( c = 4y - 4 \)

### Step 5: Apply the discriminant condition
For \( x \) to be real, the discriminant \( D \) must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the coefficients:
\[ (2y + 2)^2 - 4(1 - y)(4y - 4) \geq 0 \]

### Step 6: Simplify the discriminant
Calculating \( D \):
\[ (2y + 2)^2 = 4y^2 + 8y + 4 \]
\[ 4(1 - y)(4y - 4) = 16y - 16 - 16y^2 + 16y = -16y^2 + 32y - 16 \]
Thus, we have:
\[ 4y^2 + 8y + 4 + 16y^2 - 32y + 16 \geq 0 \]
Combining like terms:
\[ 20y^2 - 24y + 20 \geq 0 \]

### Step 7: Factor the quadratic
Dividing the entire inequality by 4:
\[ 5y^2 - 6y + 5 \geq 0 \]
Now we can find the roots using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 5 \cdot 5}}{2 \cdot 5} \]
Calculating the discriminant:
\[ 36 - 100 = -64 \]
Since the discriminant is negative, the quadratic does not cross the x-axis and is always positive.

### Step 8: Determine the interval
Since \( 5y^2 - 6y + 5 \geq 0 \) for all \( y \), we need to check the values of \( y \) that satisfy the original function.

### Step 9: Find the maximum and minimum values of \( f(x) \)
To find the range of \( f(x) \), we can analyze the limits as \( x \to \infty \) and \( x \to -\infty \):
- As \( x \to \infty \), \( f(x) \to 1 \).
- As \( x \to -\infty \), \( f(x) \to 1 \).

### Step 10: Check specific values
To find specific bounds, we can check the function at some values:
- At \( x = 0 \): \( f(0) = 1 \).
- At \( x = 1 \): \( f(1) = \frac{3}{5} \).
- At \( x = -1 \): \( f(-1) = \frac{7}{5} \).

### Conclusion
The function \( f(x) \) takes values in the interval \(\left[\frac{1}{3}, 3\right]\).