For all real values of x, the maximum value of `(1-x+x^(2))/(1+x+x^(2))` is :

Options

A. 0
B. 1
C. 3
D. `(1)/(3)`

Answer

To find the maximum value of the function \( y = \frac{1 - x + x^2}{1 + x + x^2} \), we will follow these steps:

### Step 1: Find the derivative \( \frac{dy}{dx} \)

Using the quotient rule, where \( u = 1 - x + x^2 \) and \( v = 1 + x + x^2 \), we have:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

Calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

- \( \frac{du}{dx} = -1 + 2x \)
- \( \frac{dv}{dx} = 1 + 2x \)

Substituting these into the derivative formula:

\[ \frac{dy}{dx} = \frac{(1 + x + x^2)(-1 + 2x) - (1 - x + x^2)(1 + 2x)}{(1 + x + x^2)^2} \]

### Step 2: Set the derivative to zero to find critical points

Setting the numerator equal to zero:

\[ (1 + x + x^2)(-1 + 2x) - (1 - x + x^2)(1 + 2x) = 0 \]

### Step 3: Simplify the equation

Expanding both sides:

1. Expand \( (1 + x + x^2)(-1 + 2x) \):
\[ -1 - x - x^2 + 2x + 2x^2 + 2x^3 = 2x^3 + x^2 + x - 1 \]

2. Expand \( (1 - x + x^2)(1 + 2x) \):
\[ 1 + 2x - x - 2x^2 + x^2 + 2x^3 = 2x^3 - x^2 + x + 1 \]

Setting the entire equation to zero:

\[ (2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1) = 0 \]

This simplifies to:

\[ 2x^2 - 2 = 0 \implies x^2 = 1 \implies x = \pm 1 \]

### Step 4: Determine the nature of critical points

To determine whether these points are maxima or minima, we can use the second derivative test or evaluate the first derivative around these points.

### Step 5: Evaluate the function at critical points

Calculating \( y \) at \( x = 1 \):

\[ y(1) = \frac{1 - 1 + 1^2}{1 + 1 + 1^2} = \frac{1}{3} \]

Calculating \( y \) at \( x = -1 \):

\[ y(-1) = \frac{1 - (-1) + (-1)^2}{1 + (-1) + (-1)^2} = \frac{1 + 1 + 1}{1 - 1 + 1} = \frac{3}{1} = 3 \]

### Step 6: Conclusion

The maximum value of \( y \) occurs at \( x = -1 \):

\[ \text{Maximum value} = 3 \]

### Final Answer
The maximum value of \( \frac{1 - x + x^2}{1 + x + x^2} \) for all real values of \( x \) is **3**.

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