For all real numbers x such that x \neq 0 and x \neq 1, the function f satisfies the equat… — GRE General Test

For all real numbers

x

such that

x \neq 0

and

x \neq 1

, the function

f

satisfies the equation

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

What is the value of

f(2)

Answer: 0.75

The value of

f(2)

0.75

(or

\frac{3}{4}

Evaluating the given equation at

x = 2

x = 1/2

, and

x = -1

generates a system of three linear equations:

f(2) + f(1/2) = 3

f(1/2) + f(-1) = 1.5

, and

f(-1) + f(2) = 0

. Adding all three equations yields

2(f(2) + f(1/2) + f(-1)) = 4.5

, which simplifies to

f(2) + f(1/2) + f(-1) = 2.25

. Subtracting the equation

f(1/2) + f(-1) = 1.5

from

2.25

gives

f(2) = 0.75

Determine the cyclic sequence of inputs starting at

x = 2

Evaluating the inner transformation

g(x) = \frac{x-1}{x}

yields

g(2) = \frac{1}{2}

g\left(\frac{1}{2}\right) = -1

, and

g(-1) = 2

. The input sequence cycles every 3 steps.

Identifying that

g(g(g(x))) = x

allows constructing a closed system of linear equations.

Set up a system of three linear equations using the cycle values

For

x = 2

f(2) + f\left(\frac{1}{2}\right) = 3

.
For

x = \frac{1}{2}

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

.
For

x = -1

f(-1) + f(2) = 0

Substituting each element of the cycle into the given relation produces three linear equations in three unknown function values.

Sum the equations and isolate

f(2)

Summing all three equations gives

2\left[f(2) + f\left(\frac{1}{2}\right) + f(-1)\right] = \frac{9}{2}

, so

f(2) + f\left(\frac{1}{2}\right) + f(-1) = \frac{9}{4}

. Subtracting

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

yields

f(2) = \frac{9}{4} - \frac{6}{4} = \frac{3}{4} = 0.75

Symmetric summation provides the total sum of the three function values, allowing direct elimination of the other two variables.

Solving cyclic functional equations by forming a system of linear equations

Question