If f is a real function satisfying
Question:

If $t$ is a real function satisfying $f\left(x+\frac{1}{x}\right)=x^{2}+\frac{1}{x^{2}}$ for all $x \in \mathrm{R}-\{0\}$, then write the expression for $f(x)$.

Solution:

Given:

$f\left(x+\frac{1}{x}\right)=x^{2}+\frac{1}{x^{2}}$

$=x^{2}+\frac{1}{x^{2}}+2-2$

$=\left(x+\frac{1}{x}\right)^{2}-2$

Thus,

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

Hence,

f (x)  = x2  2 , where | x | ≥  2.