Let `f(x+(1)/(x))=x^(2)+(1)/(x^(2)),(x ne 0)` then f(x) equals

Options

A. `x^(2)-` for all x
B. `x^(2)-2"for all" |x| gt 2`
C. `x^(2)-2"for all "|x|lt2`
D. none of these

Answer

To find the function \( f(x) \) given that \( f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2} \), we can follow these steps:

### Step 1: Rewrite the Right Side
We start with the equation:
\[ f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2} \]
We can rewrite the right side using the identity:
\[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \]
Thus, we have:
\[ f\left(x + \frac{1}{x}\right) = \left(x + \frac{1}{x}\right)^2 - 2 \]

### Step 2: Introduce a New Variable
Let:
\[ t = x + \frac{1}{x} \]
This implies:
\[ f(t) = t^2 - 2 \]

### Step 3: Determine the Domain
Next, we need to find the range of \( t \). The expression \( x + \frac{1}{x} \) achieves its minimum value when \( x = 1 \) or \( x = -1 \):
\[ x + \frac{1}{x} \geq 2 \quad \text{for } x > 0 \]
\[ x + \frac{1}{x} \leq -2 \quad \text{for } x < 0 \]
Thus, the values of \( t \) are:
\[ t \in (-\infty, -2] \cup [2, \infty) \]

### Step 4: Write the Final Function
From our previous steps, we can conclude that:
\[ f(t) = t^2 - 2 \quad \text{for } |t| \geq 2 \]
Substituting back for \( t \):
\[ f(x) = x^2 - 2 \quad \text{for } |x| \geq 2 \]

### Final Answer
Thus, the final answer is:
\[ f(x) = x^2 - 2 \quad \text{for } |x| \geq 2 \]

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