`f(x)=((x-2)(x-1))/(x-3), forall xgt3`. The minimum value of `f(x)` is equal to

Options

A. `3+2sqrt(2)`
B. `3+2sqrt(3)`
C. `3sqrt(2)+2`
D. `3sqrt(2)-2`

Explanation

To find the minimum value of the function \( f(x) = \frac{(x-2)(x-1)}{(x-3)} \) for \( x > 3 \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \frac{(x-2)(x-1)}{(x-3)} \]
For \( x > 3 \), we can make a substitution to simplify our calculations. Let:
\[ t = x - 3 \quad \text{(which implies } x = t + 3 \text{)} \]
Thus, we can rewrite \( f(x) \) in terms of \( t \):
\[ f(t) = \frac{((t + 3) - 2)((t + 3) - 1)}{(t + 3) - 3} = \frac{(t + 1)(t + 2)}{t} \]

### Step 2: Simplify the function
Now, simplifying \( f(t) \):
\[ f(t) = \frac{(t + 1)(t + 2)}{t} = \frac{t^2 + 3t + 2}{t} = t + 3 + \frac{2}{t} \]
So, we have:
\[ f(t) = t + 3 + \frac{2}{t} \]
for \( t > 0 \) (since \( x > 3 \) implies \( t > 0 \)).

### Step 3: Find the minimum value
To find the minimum value of \( f(t) \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. We can apply AM-GM to the terms \( t \) and \( \frac{2}{t} \):
\[ \frac{t + \frac{2}{t}}{2} \geq \sqrt{t \cdot \frac{2}{t}} = \sqrt{2} \]
This implies:
\[ t + \frac{2}{t} \geq 2\sqrt{2} \]
Adding 3 to both sides gives:
\[ f(t) = t + 3 + \frac{2}{t} \geq 2\sqrt{2} + 3 \]

### Step 4: Find when the minimum occurs
The equality in AM-GM holds when:
\[ t = \frac{2}{t} \implies t^2 = 2 \implies t = \sqrt{2} \]
Thus, the minimum value occurs at \( t = \sqrt{2} \).

### Step 5: Calculate the minimum value
Substituting \( t = \sqrt{2} \) back into \( f(t) \):
\[ f(\sqrt{2}) = \sqrt{2} + 3 + \frac{2}{\sqrt{2}} = \sqrt{2} + 3 + \sqrt{2} = 2\sqrt{2} + 3 \]

### Conclusion
Thus, the minimum value of \( f(x) \) for \( x > 3 \) is:
\[ \boxed{3 + 2\sqrt{2}} \]