Question

Let f be the function from R to R defined by $f\left(x\right)=x^2$. Find

localid="1668675673668" $$\begin{gathered} \left(a\right)f^{-1}\left(1\right) \\ \left(b\right)f^{-1}\left(x|0<x<1\right) \\ \left(c\right)f^{-1}\left(x|x>4\right) \end{gathered}$$

Short answer

The function is

$$\begin{gathered} \left(a\right)f^{-1}\left(1\right)=\{1,1\} \\ \left(b\right)f^{-1}\left(x|0<x<1\right)=x\left|\right(-1<x<1)\wedge (x\ne 0) \end{gathered}$$

$\left(c\right)f^{-1}\left(x|x>4\right)=x\left|\right(x<-2)\vee (x>2)$

Step-by-step solution

Step 1:

Union $A\cup B$: all elements that are either in A or in B.

Intersection $A\cap B$: all elements that are both in A and in B.

X is a subset of Y if every element of X is also an element of Y.

Notation:$X\subseteq Y$

Step 2:

Given: (a)

$$\begin{gathered} f:R\to R \\ f\left(x\right)=x^2 \end{gathered}$$

$f^{-1}\left\{\left(1\right)\right\}$contains all elements of R that has as image1.

$f\left(x\right)=1$

Since $f\left(x\right)=x^2$, we have

$$\begin{gathered} x^2=1 \\ {x}^2-1=0 \\ (x+1)(x-1) \end{gathered}$$

The solution of the equation are $x=1andx=-1and{f}^{-1}\left(1\right)$ thus contains -1 and 1

$f^{-1}\left(1\right)=-1,1$

Step 3:

$f^{-1}\left(\left\{x|0<x<1\right\}\right)$contains all elements of R that has as image a real number between 0 and 1.

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

Since$f\left(x\right)=x^2$, we have

$0<x^2<1$

The square$x^2$is less than 1 if x is between -1 and 1.

The square$x^2$is more than 0, when x is not equal to 0.

$(-1<x<1)\wedge (x\ne 0)$

$f^{-1}\left(\left\{x|0<x<1\right\}\right)$thus contains all real numbers between -1 and 1, except for 0.

$f^{-1}\left(\left\{x|0<x<1\right\}\right)=\left\{x\left|\right(-1<x<1)\wedge (x\ne 0)\right\}$

Step 4:

( c)$f^{-1}\left(\left\{x|0<x>4\right\}\right)$contains all elements of R that has as image a real number greater than 4.

$f\left(x\right)>4$

Since$f\left(x\right)=x^2$, we have

$x^2>4$

The square$x^2$is less than -2

The square$x^2$is more than 4, if x is more than 2 or x is less than -2

$(x<-2)\vee (x>2)$

$f^{-1}\left(\left\{x|0<x>4\right\}\right)$thus contains all real numbers smaller than -2 and larger than 2.

$\left(c\right)f^{-1}\left(\left\{x|x>4\right\}\right)=\left\{x\left|\right(x<-2)\vee (x>2)\right\}$

Hence, the solution is

$\begin{array}{r}\left(a\right)f^{-1}\left(1\right)=\{1,1\}\\ \left(b\right)f^{-1}\left(\right\{x\mid 0<x<1\left\}\right)=\{x\mid (-1<x<1)\wedge (x\ne 0\left)\right\}\\ \left(c\right)f^{-1}\left(\right\{x\mid x>4\left\}\right)=\{x\mid (x<-2)\vee (x>2\left)\right\}\end{array}$