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

Options

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

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: Analyze the function
We start with the function:
\[ f(x) = \frac{x^2 + x - 1}{x^2 - x + 1} \]

### Step 2: Rewrite the function
We can rewrite \( f(x) \) as:
\[ f(x) = \frac{x^2 - x + 1 + 2x - 2}{x^2 - x + 1} = 1 + \frac{2x - 2}{x^2 - x + 1} \]
This simplifies to:
\[ f(x) = 1 + \frac{2(x - 1)}{x^2 - x + 1} \]

### Step 3: Find critical points
To find the maximum value of \( f(x) \), we need to analyze the term \( \frac{2(x - 1)}{x^2 - x + 1} \). We will find the derivative of \( f(x) \) and set it to zero to find critical points.

Let:
\[ g(x) = \frac{2(x - 1)}{x^2 - x + 1} \]

Using the quotient rule, we find \( g'(x) \):
\[ g'(x) = \frac{(2)(x^2 - x + 1) - 2(x - 1)(2x - 1)}{(x^2 - x + 1)^2} \]

Setting \( g'(x) = 0 \) gives us the critical points.

### Step 4: Evaluate endpoints and critical points
We will evaluate \( f(x) \) at the endpoints of the interval and at any critical points found in the previous step.

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

2. Evaluate \( f(3) \):
\[ f(3) = \frac{3^2 + 3 - 1}{3^2 - 3 + 1} = \frac{9 + 3 - 1}{9 - 3 + 1} = \frac{11}{7} \]

3. Evaluate \( f\left(\frac{1}{2}\right) \) if it is a critical point:
\[ f\left(\frac{1}{2}\right) = 1 + \frac{2\left(\frac{1}{2} - 1\right)}{\left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1} = 1 + \frac{2\left(-\frac{1}{2}\right)}{\frac{1}{4} - \frac{1}{2} + 1} = 1 + \frac{-1}{\frac{3}{4}} = 1 - \frac{4}{3} = -\frac{1}{3} \]

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

### Conclusion
The largest value of \( f(x) \) in the interval \([-1, 3]\) is:
\[ \boxed{\frac{11}{7}} \]