For all non-zero real numbers x, the function f satisfies the relation 2f(x) + f\left(\fra… — GRE General Test

For all non-zero real numbers $x$ , the function $f$ satisfies the relation $2f(x) + f\left(\frac{1}{x}\right) = 3x$ . What is the value of $f(2)$ ?

The value of

f(2)

\frac{7}{2}

To solve for

f(2)

, set up a system of equations by evaluating the given relation

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

x = 2

and at

x = 1/2

. Substituting

x = 2

gives

2f(2) + f(1/2) = 6

, and substituting

x = 1/2

gives

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

. Expressing

f(1/2)

from the first equation as

6 - 2f(2)

and substituting it into the second yields

2(6 - 2f(2)) + f(2) = 3/2

, which simplifies to

-3f(2) = -21/2

, so

f(2) = 7/2

Substitute

x = 2

into the functional relation.

2f(2) + f\left(\frac{1}{2}\right) = 3(2) = 6

This yields a linear relationship involving

f(2)

and

f\left(\frac{1}{2}\right)

Substitute

x = \frac{1}{2}

into the original functional relation.

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

This provides a second independent linear equation with the same two unknown quantities,

f(2)

and

f\left(\frac{1}{2}\right)

Multiply the second equation by

2

to eliminate

f\left(\frac{1}{2}\right)

4f\left(\frac{1}{2}\right) + 2f(2) = 3

Preparing to eliminate

f(2)

f\left(\frac{1}{2}\right)

using elimination.

Solve the system of equations for

f(2)

From

2f(2) + f\left(\frac{1}{2}\right) = 6

, we get

f\left(\frac{1}{2}\right) = 6 - 2f(2)

. Substituting this into the second equation

2(6 - 2f(2)) + f(2) = \frac{3}{2} \implies 12 - 4f(2) + f(2) = \frac{3}{2} \implies -3f(2) = \frac{3}{2} - 12 = -\frac{21}{2} \implies f(2) = \frac{7}{2}

Algebraic reduction isolates

f(2)

Functional Equations and System of Equations via Variable Substitution

Question