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A real-valued function f defined for nonzero real numbers satisties  f(1/x)+1/xf(-x) = 2x. What is the value of f(2)?

 Oct 16, 2018
 #1
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\(f\left(\dfrac 1 x\right)+\dfrac 1 x f(-x) = 2x ~;\text{now let }x \to -\dfrac 1 x \\ f(-x) -x f\left(\dfrac 1 x\right) = -\dfrac 2 x~; \text{and multiply by }\dfrac 1 x \\ \dfrac 1 x f(-x) - f\left(\dfrac 1 x\right) = -\dfrac{2}{x^2}\)

 

\(\text{now add this to the original equation }\\ \dfrac 2 x f(-x) = 2x - \dfrac{2}{x^2} = \dfrac{2x^3-2}{x^2} \\ f(-x) = \dfrac{x^3-1}{x} = x^2 - \dfrac 1 x\\ f(x) = x^2 + \dfrac 1 x\)

 

\(f(2) = 2^2 +\dfrac 1 2 = \dfrac 9 2\)

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 Oct 17, 2018

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