Question

A function f satisfies the condition $f(x+\frac{1}{x})=x^2+\frac{1}{x^2},x\ne 0.$ Determine f(2)

Answer 1

$f(x+\frac{1}{x})=x^2=\frac{1}{x^2}$
$=(x+\frac{1}{x}{)}^2-2$
Therefore
$f(x+\frac{1}{x}{)}^2=(x+\frac{1}{x}{)}^2-2$
Therefore
$f\left(x\right)=x^2-2$
Hence
$f\left(2\right)=2^2-2$
$=4-2$
$=2$.