If f is a function defined on (0, 1) by f(x) | vedclass

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(A) Given $f(x) = \min \{x - [x], -x - [-x]\}$ for $x \in (0, 1)$.Since $x \in (0, 1)$,$[x] = 0$.Also,for $x \in (0, 1)$,$-x \in (-1, 0)$,so $[-x] = -

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$x$

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(A) Given $f(x) = \min \{x - [x], -x - [-x]\}$ for $x \in (0, 1)$.Since $x \in (0, 1)$,$[x] = 0$.Also,for $x \in (0, 1)$,$-x \in (-1, 0)$,so $[-x] = -1$.Substituting these values into the function:$f(x) = \min \{x - 0, -x - (-1)\} = \min \{x, 1 - x\}$.Now,we evaluate the composition $(f \circ f \circ f \circ f)(x)$:If $x \in (0, 1/2]$,then $x \le 1 - x$,so $f(x) = x$.Then $f(f(x)) = f(x) = x$,and so on. Thus $(f \circ f \circ f \circ f)(x) = x$.If $x \in (1/2, 1)$,then $1 - x Then $f(f(x)) = f(1 - x)$. Since $1 - x \in (0, 1/2)$,$f(1 - x) = 1 - x$.Wait,let us re-evaluate: $f(1-x) = \min \{1-x, 1-(1-x)\} = \min \{1-x, x\} = 1-x$ if $1-x  1/2$).Actually,for $x \in (0, 1)$,$f(x) = \min \{x, 1-x\}$.$f(f(x)) = f(\min \{x, 1-x\})$.If $x \in (0, 1/2]$,$f(x) = x$,so $f(f(x)) = f(x) = x$.If $x \in (1/2, 1)$,$f(x) = 1-x$,so $f(f(x)) = f(1-x) = \min \{1-x, 1-(1-x)\} = \min \{1-x, x\} = 1-x$.Thus,$(f \circ f)(x) = f(x)$.Consequently,$(f \circ f \circ f \circ f)(x) = f(f(x)) = f(x)$.

If $f$ is a function defined on $(0, 1)$ by $f(x) = \min \{x - [x], -x - [-x]\}$,then $(f \circ f \circ f \circ f)(x)$ is equal to ($[.]$ denotes the greatest integer function).

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