Quadratic Functional Equation

Algebra Level 3

If f ( x + 1 x ) = x 2 + 1 x 2 + 1 x f\left(\dfrac{x+1}{x}\right) = \dfrac{x^2+1}{x^2} + \dfrac{1}{x} , then which is the following is a possible representation of f ( x ) f(x) ?

x 2 + 2 x + 2 x^2+2x+2 x 2 − 2 x + 2 x^2-2x+2 x 2 + x + 1 x^2+x+1 x 2 − x + 1 x^2-x+1

Fahim Shahriar Shakkhor by Fahim Shahriar Shakkhor , 24, Bangladesh

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2 solutions

Let t = x + 1 x t = \dfrac{x+1}{x} . Then x = 1 t − 1 x=\dfrac{1}{t-1} .

By replacing we get,

f ( t ) = ( 1 t − 1 ) 2 + 1 ( 1 t − 1 ) 2 + 1 1 t − 1 = 1 + t 2 − 2 t + 1 + t − 1 = t 2 − t + 1 f(t) = \dfrac{\left(\dfrac{1}{t-1}\right)^2+1}{\left(\dfrac{1}{t-1}\right)^2} + \dfrac{1}{\dfrac{1}{t-1}} = 1 + t^2-2t+1+t-1=\boxed{t^2-t+1}

Here t t is a dummy variable and we can rename it as x x .

Thus f ( x ) = x 2 − x + 1 f(x) = \boxed{x^2-x+1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Arian Tashakkor
May 7, 2015

f ( x + 1 x ) = ( x + 1 x ) 2 − ( x + 1 x ) + 1 → f ( x ) = x 2 − x + 1 f(\frac{x+1}{x}) = (\frac{x+1}{x})^2 -(\frac{x+1}{x}) +1 \rightarrow f(x) = x^2-x+1

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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