`f(x)=x^2+1/(x^2)-6x-6/x+2` then min value of f(x)

Options

A. `-2`
B. `-8`
C. `-9`
D. `-12`

Answer

To find the minimum value of the function \( f(x) = x^2 + \frac{1}{x^2} - 6x - \frac{6}{x} + 2 \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = x^2 + \frac{1}{x^2} - 6x - \frac{6}{x} + 2 \]

### Step 2: Combine terms
Notice that we can combine \( x^2 + \frac{1}{x^2} \) and \( -6x - \frac{6}{x} \):
\[ f(x) = \left(x^2 - 6x + 2\right) + \left(\frac{1}{x^2} - \frac{6}{x}\right) \]

### Step 3: Rewrite \( x^2 + \frac{1}{x^2} \)
We can express \( x^2 + \frac{1}{x^2} \) in terms of \( t = x + \frac{1}{x} \):
\[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = t^2 - 2 \]

### Step 4: Substitute \( t \)
Now, we can rewrite \( f(x) \):
\[ f(x) = (t^2 - 2) - 6(x + \frac{1}{x}) + 2 \]
This simplifies to:
\[ f(x) = t^2 - 6t \]

### Step 5: Find the minimum value of \( t^2 - 6t \)
To find the minimum value, we can complete the square:
\[ f(t) = t^2 - 6t = (t - 3)^2 - 9 \]

### Step 6: Determine the minimum value
The expression \( (t - 3)^2 \) is minimized when \( t = 3 \). Therefore, the minimum value of \( f(t) \) occurs at:
\[ f(3) = (3 - 3)^2 - 9 = 0 - 9 = -9 \]

### Step 7: Check the domain of \( t \)
Since \( t = x + \frac{1}{x} \) is always greater than or equal to 2, and since 3 is greater than 2, this value is valid.

### Final Answer
Thus, the minimum value of \( f(x) \) is:
\[ \boxed{-9} \]

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