The max. and min. value of `(x^2 + x+1)/(x^2 - x + 1)` are

Options

A. (2,1)
B. `(3,1/3)`
C. `(1,0)`
D. `(3,1)`

Answer

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

### Step 1: Differentiate the function
We need to find the derivative of \( f(x) \) and set it to zero to find critical points. We will use the quotient rule for differentiation.

The quotient rule states that if \( f(x) = \frac{g(x)}{h(x)} \), then:
\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \]

Here, \( g(x) = x^2 + x + 1 \) and \( h(x) = x^2 - x + 1 \).

Calculating the derivatives:
- \( g'(x) = 2x + 1 \)
- \( h'(x) = 2x - 1 \)

Now applying the quotient rule:
\[ f'(x) = \frac{(2x + 1)(x^2 - x + 1) - (x^2 + x + 1)(2x - 1)}{(x^2 - x + 1)^2} \]

### Step 2: Set the derivative equal to zero
We set the numerator equal to zero:
\[ (2x + 1)(x^2 - x + 1) - (x^2 + x + 1)(2x - 1) = 0 \]

### Step 3: Simplify the equation
Expanding both terms:
1. \( (2x + 1)(x^2 - x + 1) = 2x^3 - 2x^2 + 2x + x^2 - x + 1 = 2x^3 - x^2 + x + 1 \)
2. \( (x^2 + x + 1)(2x - 1) = 2x^3 + 2x^2 + 2x - x^2 - x - 1 = 2x^3 + x^2 + x - 1 \)

Now, substituting back:
\[ (2x^3 - x^2 + x + 1) - (2x^3 + x^2 + x - 1) = 0 \]
This simplifies to:
\[ -2x^2 + 2 = 0 \implies x^2 = 1 \implies x = \pm 1 \]

### Step 4: Evaluate the function at critical points
Now we evaluate \( f(x) \) at the critical points \( x = 1 \) and \( x = -1 \).

1. For \( x = 1 \):
\[ f(1) = \frac{1^2 + 1 + 1}{1^2 - 1 + 1} = \frac{3}{1} = 3 \]

2. For \( x = -1 \):
\[ f(-1) = \frac{(-1)^2 + (-1) + 1}{(-1)^2 - (-1) + 1} = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3} \]

### Step 5: Determine maximum and minimum values
From the evaluations:
- \( f(1) = 3 \) (maximum value)
- \( f(-1) = \frac{1}{3} \) (minimum value)

### Conclusion
The maximum value of \( f(x) \) is \( 3 \) and the minimum value is \( \frac{1}{3} \).

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