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Chapter 1: Problem 105

Given \(f(x)=1+x, \quad-1 \leq x \leq 0\) \(=-x, \quad 00\) ix. \(f(x)\) is injective function.

Short Answer

Expert verified
All determined functions are as follows: i. \(f_{1}(x)=-x, \quad-1 \leq x<0; \quad=x-1, \quad 0 \leq x<1; \quad=1, \quad x=1\). ii. \(f_{2}(x)=-x, \quad-1 \leq x \leq 0; \quad=x-1, \quad 00\). None of the above functions is injective.

Step by step solution

01

Determine \(f_{1}(x)=f(g(x))\)

To determine \(f_{1}(x)=f(g(x))\), plug \(g(x)\) into \(f(x)\). This results in \(f_{1}(x)=-x\) for \(-1 \leq x<0\) and \(f_{1}(x)=x-1\) for \(0 \leq x<1\) and \(f_{1}(x)=1\) for \(x=1\).
02

Determine \(f_{2}(x)=g(f(x))\)

Similarly, to determine \(f_{2}(x)=g(f(x))\), plug \(f(x)\) into \(g(x)\). This gives \(f_{2}(x)=-x\) for \(-1 \leq x \leq 0\) and \(f_{2}(x)=x-1\) for \(0<x \leq 1\).
03

Determine \(f_{3}(x)=f_{1}\left(f_{2}(x)\right)\)

To determine \(f_{3}(x)=f_{1}\left(f_{2}(x)\right)\), plug \(f_{2}(x)\) into \(f_{1}(x)\) for each interval. This gives \(f_{3}(x)=1\) for \(x=-1\), \(f_{3}(x)=-x-1\) for \(-1<x \leq 0\), \(f_{3}(x)=1-x\) for \(0<x<1\), and \(f_{3}(x)=-1\) for \(x=1\).
04

Determine \(f_{4}(x)=f_{2}\left(f_{1}(x)\right)\)

To determine \(f_{4}(x)=f_{2}\left(f_{1}(x)\right)\), plug \(f_{1}(x)\) into \(f_{2}(x)\) for each interval. This results in \(f_{4}(x)=-x-1\) for \(-1 \leq x<0\) and \(f_{4}(x)=1-x\) for \(0 \leq x \leq 1\).
05

Determine \(f_{5}(x)=f(|x|)\)

Computing \(f_{5}(x)=f(|x|)\), we recall that \(|x|\) equals \(x\) if \(x \geq 0\) and \(-x\) else. Plugging this into \(f(x)\) yields \(f_{5}(x)=x\) for \(-1 \leq x<0\), \(f_{5}(x)=1\) for \(x=0\), and \(f_{5}(x)=-x\) for \(0<x \leq 1\).
06

Determine \(f_{6}(x)=|f(x)|\)

By computing \(f_{6}(x)=|f(x)|\), we take the absolute value of each part of the piecewise function \(f(x)\). The results are \(f_{6}(x)=1+x\) for \(-1 \leq x \leq 0\) and \(f_{6}(x)=x\) for \(0<x \leq 1\).
07

Determine \(f_{7}(x)=\operatorname{sgn}(f(x))\)

To figure out \(f_{7}(x)=\operatorname{sgn}(f(x))\), take the signum function of each part of \(f(x)\). When we apply the signum function, \(f_{7}(x)=0\) for \(x=-1\), \(f_{7}(x)=1\) for \(-1<x \leq 0\), and \(f_{7}(x)=-1\) for \(0<x \leq 1\).
08

Determine \(f_{8}(x)=f(\operatorname{sgn}(x))\)

Lastly, to determine \(f_{8}(x)=f(\operatorname{sgn} x)\), substitute the signum of \(x\) into \(f(x)\). The outcome is \(f_{8}(x)=0\) for \(x<0\), \(f_{8}(x)=1\) for \(x=0\), and \(f_{8}(x)=-1\) for \(x>0\).
09

Determine Which Functions are Injective

An injective function, also called one-to-one, is a function for which no two different inputs give the same output. By observing each combined function, we can see that the function \(f(x)\), consists of multiple intervals where the same output can be achieved by different inputs, hence it is not an injective function. But if also functions like \(f_{1}(x)\) are considered, they also show this property, hence also they are not injective. In fact, all combined functions are not injective.

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