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
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