Functions:

Algebra Level 3

$\large f(x) + f \left(\frac 1{1-x}\right) = x$

Find a function that satisfies the equation above.

\frac {x^3 - x + 1}{2x(x-1)}

\frac {x^2 - x - 1}{1-x}

\frac {15x-10}{x-1}

x-\frac 1x

2 solutions

Christian Daang
Oct 12, 2014

Solution:

1) f(x) + f(1/(1-x)) = x or f(x) = x - f(1/(1-x))

2) f(1/(1-x)) + f((x-1)/x) = (1/(1-x)) or f(1/(1-x)) = 1/(1-x) - f((x-1)/x)

3) f((x-1)/x) + f(x) = (x-1)/x or f((x-1)/x) = (x-1)/x - f(x)

f((x-1)/x) = (x-1)/x - f(x)

= (x-1)/x - x + f(1/(1-x))

= (x-1)/x - x + 1/(1-x) - f((x-1)/x)

2f((x-1)/x) = {(x-1)/x + 1/(1-x)} - x

f((x-1)/x) = (x^3 - 2x^2 + 3x - 1)/(2x*(1-x))

f(x) = (x-1)/x - f((x-1)/x)

= (x-1)/x - (x^3 - 2x^2 + 3x - 1)/(2x*(1-x))

there fore,

f(x) = (x^3 - x + 1)/(2x*(x-1)) ; where in, x =/= (0,1)...

u have given f(x) - f(1/1-x) = x... u have given "-" sign not "+" between f(x) and f(1/1-x)

rahul singh - 4 years, 1 month ago

Barr Shiv
Dec 10, 2018

that is the only function that is not defined at x=0 and 1