Let `f (x) =(x^(2) +x-1)/(x ^(2) - x+1),` then the largest vlaue of `f (x) AA x in [-1, 3]` is:

Options

A. `3/5`
B. `5/3`
C. `1`
D. `4/3`

Answer

To find the largest value of the function \( f(x) = \frac{x^2 + x - 1}{x^2 - x + 1} \) for \( x \) in the interval \([-1, 3]\), we will follow these steps:

### Step 1: Find the derivative of \( f(x) \)
We will use the quotient rule to find the derivative \( f'(x) \):
\[ f'(x) = \frac{(u'v - uv')}{v^2} \]
where \( u = x^2 + x - 1 \) and \( v = x^2 - x + 1 \).

Calculating \( u' \) and \( v' \):
- \( u' = 2x + 1 \)
- \( v' = 2x - 1 \)

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 to zero
To find the critical points, we set the numerator equal to zero:
\[ (2x + 1)(x^2 - x + 1) - (x^2 + x - 1)(2x - 1) = 0 \]

### Step 3: Solve the equation
Expanding both terms and simplifying:
\[ 2x^3 - 2x^2 + 2x + x^2 - x + 1 - (2x^3 - 2x^2 + 2x - x^2 - x + 1) = 0 \]
This simplifies to:
\[ -2x^2 + 4x = 0 \]
Factoring gives:
\[ 2x(2 - x) = 0 \]
Thus, the critical points are \( x = 0 \) and \( x = 2 \).

### Step 4: Evaluate \( f(x) \) at critical points and endpoints
We need to evaluate \( f(x) \) at the critical points and the endpoints of the interval \([-1, 3]\):
1. \( f(-1) = \frac{(-1)^2 + (-1) - 1}{(-1)^2 - (-1) + 1} = \frac{1 - 1 - 1}{1 + 1 + 1} = \frac{-1}{3} = -\frac{1}{3} \)
2. \( f(0) = \frac{0^2 + 0 - 1}{0^2 - 0 + 1} = \frac{-1}{1} = -1 \)
3. \( f(2) = \frac{2^2 + 2 - 1}{2^2 - 2 + 1} = \frac{4 + 2 - 1}{4 - 2 + 1} = \frac{5}{3} \)
4. \( f(3) = \frac{3^2 + 3 - 1}{3^2 - 3 + 1} = \frac{9 + 3 - 1}{9 - 3 + 1} = \frac{11}{7} \)

### Step 5: Compare values
Now we compare the values:
- \( f(-1) = -\frac{1}{3} \)
- \( f(0) = -1 \)
- \( f(2) = \frac{5}{3} \)
- \( f(3) = \frac{11}{7} \)

The largest value among these is \( f(2) = \frac{5}{3} \).

### Conclusion
Thus, the largest value of \( f(x) \) in the interval \([-1, 3]\) is:
\[ \boxed{\frac{5}{3}} \]