Function(function)

Algebra Level 2

Let $f(x)$ be defined on $[0,1]$ by the rule $f(x)=\begin{cases}x & \mbox{if }x\in\mathbb{Q} \\ 1-x & \mbox{if }x\notin\mathbb{Q}\end{cases}$ Then for all $x\in[0,1]$ , $f(f(x))$ is

x

Constant

1+x

None of the others

1 solution

Tom Engelsman
May 22, 2021

If $x \in \mathbb{Q}$ , then $f(f(x)) = x$ . If $x \notin \mathbb{Q}$ , then $f(f(x)) = 1-(1-x) = x$ . Thus $\boxed{f(f(x)) =x}$ for all $x \in [0,1].$

Function(function)

1 solution

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