For all non-zero real numbers x, the function f satisfies the functional equation f(x) + 2… — GRE General Test

For all non-zero real numbers

x

, the function

f

satisfies the functional equation

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

Which of the following statements must be true for all $x \neq 0$ ? Select all such statements.

$f(-1) = 3$ Answer
B
$f(x) = f\left(-\frac{1}{x}\right)$ for all $x \neq 0$
$f(x) \leq -2\sqrt{2}$ for all $x > 0$ Answer
D
$f(x) + f(-x) = 6x$ for all $x \neq 0$
E
$f(2) = 3$

Answer

The statements stating that

f(-1) = 3

and that

f(x) \leq -2\sqrt{2}

for all

x > 0

are the correct choices.

Solving the functional equation yields

f(x) = -x - \frac{2}{x}

. Substituting

x = -1

gives

f(-1) = 3

, making the statement asserting

f(-1) = 3

correct. Furthermore, for any positive real number

x

, the sum

x + \frac{2}{x} \geq 2\sqrt{2}

by the AM-GM inequality, so

f(x) = -\left(x + \frac{2}{x}\right) \leq -2\sqrt{2}

, making the statement asserting

f(x) \leq -2\sqrt{2}

for all

x > 0

correct.

Step-by-Step Solution

Set up a system of functional equations by substituting

x \to -\frac{1}{x}

Replacing

x

with

-\frac{1}{x}

in the original equation

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

gives

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

This creates a system of two linear algebraic equations with two unknowns:

f(x)

and

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

Solve the system of equations for

f(x)

Multiply the second equation by

2

to obtain

4f(x) + 2f\left(-\frac{1}{x}\right) = -\frac{6}{x}

. Subtracting the first equation

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

from this gives

3f(x) = -3x - \frac{6}{x}

, so

f(x) = -x - \frac{2}{x}

Eliminating the composite term

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

yields an explicit expression for

f(x)

Evaluate

f(-1)

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

This confirms that the statement asserting

f(-1) = 3

is correct.

Analyze the maximum value of

f(x)

for

x > 0

using the AM-GM inequality.

For

x > 0

x + \frac{2}{x} \geq 2\sqrt{x \cdot \frac{2}{x}} = 2\sqrt{2}

. Therefore,

f(x) = -\left(x + \frac{2}{x}\right) \leq -2\sqrt{2}

This confirms that the statement asserting

f(x) \leq -2\sqrt{2}

for all

x > 0

is correct.

Key Concept

Solving functional equations via variable substitution and analyzing function bounds via the AM-GM inequality.

Estimated Time:2m 30s

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