let $f( x + \frac{ 1}{ x } ) = x ^ 2 + \frac{ 1}{ x ^ 2} $ then $f(x)$ equals

Question

let $f( x + \frac{ 1}{ x } ) = x ^ 2 + \frac{ 1}{ x ^ 2} $ then $f(x)$ equals

options are

$( A ) x ^ 2 - 2$

$( B ) x ^ 2 - 1$

$( C ) x ^ 2$

I don't know how to solve this type of problems

Answer 1

The option is A. Just substitute $f(x+\frac 1x)$ in option A and verify the outcome

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

So this outcome is only possible if $f(x)=x^2-2$

Answer 2

Hint. Note that $(a+b)^2 =a^2 +2ab +b^2$. Now let $a=x$ and $b=1/x$. Then relate $f(x+\frac1x)$ with $(x+\frac1x)^2$ somehow.

Answer 3

$$f(x+\frac1x)=x^2+\frac{1}{x^2}$$

Let's run through the options.

First, let's try $A$. $f(x)=x^2-2$

Thus $$f(x+\frac 1x)=(x+\frac 1x)^2-2$$ Evaluating the $RHS$ we get: $$(x+\frac1x)^2-2=x^2+\frac{1}{x^2}+2-2=x^2+\frac{1}{x^2}$$ as required. No need, therefore, to check $B$ and $C$.

Answer 4

Lets to this the mathematical way. Let $a=x+\frac 1x$. Simply $f(a) = a^2-2$. Now by obvious bijection and no known constraints, one to one mapping will obviously show $f(x) = x^2-2$.

So choose options A.

This seems a more complete answer, as OP requires if no options were provided.