Question

Show \(g(x) = |x|\). Find

\(\begin{array}{l}(a){g^{ - 1}}(\{ 0\} )\\(b){g^{ - 1}}(\{ - 1,0,1\} )\\(c){g^{ - 1}}(\{ x|0 < x < 1\} )\end{array}\)

Short answer

The function is

\(\begin{array}{l}(a){g^{ - 1}}(0) = \{ x|0 \le x < 1\} \\(b){g^{ - 1}}(\{ - 1,0,1\} ) = \{ x| - 1 \le x < 2\} \\(c){g^{ - 1}}(\{ x|0 < x < 1\} ) = \emptyset \end{array}\)

Step-by-step solution

Step 1:

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

Step 2:

Given: (a)

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

\({g^{ - 1}}(\{ 0\} )\)contains all elements of R that has as image0.

\(g(x) = 0\)

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

\(\left\lfloor x \right\rfloor = 0\)

\(0 \le x < 1\)

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

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

Step 3:Part (b)

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

\(g(x) = - 1 \vee g(x) = 0 \vee g(x) = 1\)

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

\(\left\lfloor x \right\rfloor = - 1 \vee \left\lfloor x \right\rfloor = 0 \vee \left\lfloor x \right\rfloor = 1\)

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

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

Step 4:Part ( c)

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

\(0 < g(x) < 1\)

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

\(0 < \left\lfloor x \right\rfloor < 1\)

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

\({g^{ - 1}}(\{ x|0 < x < 1\} ) = \emptyset \)

Hence, the solution is

\(\begin{array}{l}(a){g^{ - 1}}(0) = \{ x|0 \le x < 1\} \\(b){g^{ - 1}}(\{ - 1,0,1\} ) = \{ x| - 1 \le x < 2\} \\(c){g^{ - 1}}(\{ x|0 < x < 1\} ) = \emptyset \end{array}\)