For all real numbers x \neq 0 and x \neq 1, the function f is defined by f(x) = \frac{x - … — GRE General Test

For all real numbers $x \neq 0$ and $x \neq 1$ , the function $f$ is defined by $f(x) = \frac{x - 1}{x}$ . The composite function notation $f^n(x)$ is defined by $f^1(x) = f(x)$ and $f^n(x) = f\left(f^{n-1}(x)\right)$ for all integers $n \ge 2$ . What is the value of $f^{2026}(5)$ ?

A
$-\frac{5}{4}$
B
$-\frac{1}{4}$
$\frac{4}{5}$ Answer
D
$\frac{5}{4}$
E
$5$

Answer

\frac{4}{5}

Evaluating successive compositions of

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

starting at

x = 5

yields

f^1(5) = \frac{4}{5}

f^2(5) = -\frac{1}{4}

, and

f^3(5) = 5

. This produces a repeating sequence of period 3. Since

2026 \equiv 1 \pmod 3

f^{2026}(5)

equals

f^1(5) = \frac{4}{5}

Step-by-Step Solution

Evaluate the first iteration

f^1(5)

f^1(5) = f(5) = \frac{5 - 1}{5} = \frac{4}{5}

Substitute

x = 5

into the definition of

f(x)

Evaluate the second iteration

f^2(5) = f(f^1(5))

f^2(5) = f\left(\frac{4}{5}\right) = \frac{\frac{4}{5} - 1}{\frac{4}{5}} = \frac{-\frac{1}{5}}{\frac{4}{5}} = -\frac{1}{4}

Substitute

x = \frac{4}{5}

into the function definition.

Evaluate the third iteration

f^3(5) = f(f^2(5))

f^3(5) = f\left(-\frac{1}{4}\right) = \frac{-\frac{1}{4} - 1}{-\frac{1}{4}} = \frac{-\frac{5}{4}}{-\frac{1}{4}} = 5

Substitute

x = -\frac{1}{4}

into the function definition.

Identify the periodic pattern and evaluate

f^{2026}(5)

Since

f^3(5) = 5

, the function values repeat in a cycle of length 3:

(\frac{4}{5}, -\frac{1}{4}, 5)

. Dividing 2026 by 3 yields

2026 = 3 \times 675 + 1

, giving a remainder of 1. Therefore,

f^{2026}(5) = f^1(5) = \frac{4}{5}

The remainder determines the equivalent position in the 3-element repeating sequence.

Key Concept

Nested Function Composition and Periodicity

Question

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