Question

A suitable restriction on the domain of the function $f\left(x\right)={(x-1)}^2$to make it one-to-one would be _____ .

Short answer

A suitable restriction on the domain of the function f (x) = (x - 1)² to make it one-to-one would be $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{1}\mathbf{]}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{[}\mathbf{1}\mathbf{,}\mathbf{\infty }\mathbf{)}$ .

Step-by-step solution

Step 1. Given Information:

Given statement:A suitable restriction on the domain of the function f (x) = (x - 1)² to make it one-to-one would be _____ .

We want to fill the gap in the given statement.

Step 2. Solution:

A function is one to one if no two elements in the domain of f correspond to the same element in the range of f.

We notice that following:

$$\begin{gathered} f\left(0\right)={(0-1)}^2=1 \\ f\left(2\right)={(2-1)}^2=1 \end{gathered}$$

We also have:

$$\begin{gathered} f(-1)={(-1-1)}^2=4 \\ f\left(3\right)={(3-1)}^2=4 \end{gathered}$$

Hence, the given functions is not one to one.

We notice that when x is less than 1, we will end up with two elements corresponding to the same element in the range of f.

Therefore, a suitable restriction on domain of the function $f\left(x\right)={(x-1)}^2$would be $x\ge 1$

Hence, the domain of the function would be $[1,\infty )$

On the other hand, we notice that when x is greater than 1, we will end up with two elements corresponding to the same element in the range of f.

Therefore, a suitable restriction on domain of the function $f\left(x\right)={(x-1)}^2$would be $x\le 1$

Hence, the domain of the function would be$(-\infty ,1]$