Given that \( f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2} \), we need to find \( f(3) \).
Solution:
First, express \( x^2 + \frac{1}{x^2} \) in terms of \( \left(x + \frac{1}{x}\right) \):
\( \begin{aligned} \left(x + \frac{1}{x}\right)^2 &= x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} \\ &= x^2 + 2 + \frac{1}{x^2} \\ \end{aligned} \)Subtract 2 from both sides:
\( \begin{aligned} x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 – 2 \\ \end{aligned} \)Thus, the function can be rewritten as:
\( \begin{aligned} f\left(x + \frac{1}{x}\right) = \left(x + \frac{1}{x}\right)^2 – 2 \\ \end{aligned} \)Let \( u = x + \frac{1}{x} \). Then: \( f(u) = u^2 – 2 \).
Therefore, \( f(3) = 3^2 – 2 = 9 – 2 = 7 \)