Find the maximum value of the function `1//(x^(2)-3x+2)`.

Options

A. `11//4`
B. `1//4`
C. 0
D. None of these

Answer

To find the maximum value of the function \( f(x) = \frac{1}{x^2 - 3x + 2} \), we need to determine when the denominator \( g(x) = x^2 - 3x + 2 \) is at its minimum, since the maximum of \( f(x) \) occurs when \( g(x) \) is minimized.

### Step 1: Find the function to minimize
We start with the function:
\[ g(x) = x^2 - 3x + 2 \]

### Step 2: Differentiate the function
Next, we differentiate \( g(x) \) with respect to \( x \):
\[ g'(x) = 2x - 3 \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ 2x - 3 = 0 \]
Solving for \( x \):
\[ 2x = 3 \implies x = \frac{3}{2} \]

### Step 4: Determine if it's a minimum or maximum
To confirm whether this critical point is a minimum or maximum, we take the second derivative:
\[ g''(x) = 2 \]
Since \( g''(x) > 0 \), this indicates that \( g(x) \) has a minimum at \( x = \frac{3}{2} \).

### Step 5: Find the minimum value of \( g(x) \)
Now we substitute \( x = \frac{3}{2} \) back into \( g(x) \) to find the minimum value:
\[ g\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 2 \]
Calculating this:
\[ = \frac{9}{4} - \frac{9}{2} + 2 \]
Converting \( \frac{9}{2} \) to quarters:
\[ = \frac{9}{4} - \frac{18}{4} + \frac{8}{4} = \frac{9 - 18 + 8}{4} = \frac{-1}{4} \]

### Step 6: Find the maximum value of \( f(x) \)
Since \( f(x) = \frac{1}{g(x)} \), we find the maximum value of \( f(x) \) when \( g(x) \) is at its minimum:
\[ f\left(\frac{3}{2}\right) = \frac{1}{g\left(\frac{3}{2}\right)} = \frac{1}{-\frac{1}{4}} = -4 \]

### Conclusion
Thus, the maximum value of the function \( f(x) = \frac{1}{x^2 - 3x + 2} \) is:
\[ \boxed{-4} \]