An algebra problem by Wildan Bagus Wicaksono

Algebra Level 2

Given the function f f so that f ( x ) + 2 f ( 1 x ) = 3 x f(x)+2f\left( \frac { 1 }{ x } \right) =3x , for every x ≠ 0 x\neq 0 . Find the sum of all values of x x that satisfy f ( x ) = f ( − x ) f (x) = f (-x) .


The answer is 0.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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1 solution

f ( x ) + 2 f ( 1 x ) = 3 x . . . ( 1 f ( 1 x ) + 2 f ( x ) = 3 x f ( 1 x ) = 3 x − 2 f ( x ) . . . ( 2 f(x)+2f\left( \frac { 1 }{ x } \right) =3x...(1\\ f\left( \frac { 1 }{ x } \right) +2f(x)=\frac { 3 }{ x } \\ f\left( \frac { 1 }{ x } \right) =\frac { 3 }{ x } -2f(x)...(2\\

Substitution of equation ( 2 (2 into equation ( 1 (1 .

f ( x ) + 2 { 3 x − 2 f ( x ) } = 3 x f ( x ) + 6 x − 4 f ( x ) = 3 x − 3 f ( x ) = 3 x − 6 x f ( x ) = 2 x − x f(x)+2\left\{ \frac { 3 }{ x } -2f(x) \right\} =3x\\ f(x)+\frac { 6 }{ x } -4f(x)=3x\\ -3f(x)=3x-\frac { 6 }{ x } \\ f(x)=\frac { 2 }{ x } -x

Given f ( x ) = f ( − x ) f(x)=f(-x) .

2 x − x = 2 − x − ( − x ) 2 x − x = x − 2 x 4 x = 2 x 2 x = x 2 = x 2 x 1 = 2 x 2 = − 2 \frac { 2 }{ x } -x=\frac { 2 }{ -x } -(-x)\\ \frac { 2 }{ x } -x=x-\frac { 2 }{ x } \\ \frac { 4 }{ x } =2x\\ \frac { 2 }{ x } =x\\ 2={ x }^{ 2 }\\ { x }_{ 1 }=\sqrt { 2 } \\ { x }_{ 2 }=-\sqrt { 2 }

Thus, the sum of all values of x x is 2 + ( − 2 ) = 0 \sqrt { 2 } +(-\sqrt { 2 } )=0 .

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