Question

Show $g\left(x\right)=\left|x\right|$. Find

$\begin{array}{r}\left(a\right)g^{-1}\left(\right\{0\left\}\right)\\ \left(b\right)g^{-1}\left(\right\{-1,0,1\left\}\right)\\ \left(c\right)g^{-1}\left(\right\{x\mid 0<x<1\left\}\right)\end{array}$

Short answer

The function is

$\begin{array}{r}\left(a\right)g^{-1}\left(0\right)=\{x\mid 0\le x<1\}\\ \left(b\right)g^{-1}\left(\right\{-1,0,1\left\}\right)=\{x\mid -1\le x<2\}\\ \left(c\right)g^{-1}\left(\right\{x\mid 0<x<1\left\}\right)=\varnothing \end{array}$

Step-by-step solution

Step 1:

Floor function x: largest integer that is less than or equal to x.

Step 2:

Given: (a)

$g\left(x\right)=\lfloor x\rfloor$

role="math" localid="1668676108237" $g^{-1}\left(\left\{0\right\}\right)$contains all elements of R that has as image0.

$g\left(x\right)=0$

Since $g\left(x\right)=\lfloor x\rfloor$,

$$\begin{gathered} \lfloor x\rfloor =0 \\ 0\le x<1 \end{gathered}$$

$g^{-1}\left(\left\{0\right\}\right)$thus contains all real numbers between 0 (including) and 1 (including)

$g^{-1}\left(\left\{0\right\}\right)=\left\{x|0\le x<1\right\}$

Step 3:Part (b)

$f^{-1}\left(\left\{-1,0,1\right\}\right)$contains all elements of R that has as image -1,0 or 1.

$g\left(x\right)=-1\vee g\left(x\right)=0\vee g\left(x\right)=1$

Since $g\left(x\right)=\lfloor x\rfloor$,

$\lfloor x\rfloor =-1\vee \lfloor x\rfloor =0\vee \lfloor x\rfloor =1$

$g^{-1}\left(\left\{-1,0,1\right\}\right)$thus contains all real numbers between -1 (including) and 2 (including)

$g^{-1}\left(\left\{-1,0,1\right\}\right)=\left\{x\left|\right(-1\le x<0)\vee (0\le x<1\right\}$

Step 4:Part ( c)

$g^{-1}\left(\left\{x|0<x<1\right\}\right)$contains all elements of R that has as image 0

$0<g\left(x\right)<1$

Since$g\left(x\right)=\lfloor x\rfloor$,

$0<\lfloor x\rfloor <1$

$g^{-1}\left(\left\{x|0<x<1\right\}\right)$then does not contain any element so$g^{-1}\left(\left\{x|0<x<1\right\}\right)$is then the empty set:

$g^{-1}\left(\left\{x|0<x<1\right\}\right)=\varnothing$

Hence, the solution is

$\begin{array}{r}\left(a\right)g^{-1}\left(0\right)=\{x\mid 0\le x<1\}\\ \left(b\right)g^{-1}\left(\right\{-1,0,1\left\}\right)=\{x\mid -1\le x<2\}\\ \left(c\right)g^{-1}\left(\right\{x\mid 0<x<1\left\}\right)=\varnothing \end{array}$