Chapter 5.2, Problem 18AYU

In Problems 13-20, determine whether the function is one-to-one.

$\left\{\left(1,2\right),\left(2,8\right),\left(3,18\right),\left(4,32\right)\right\}$

To determine

To find: Whether the function is one-to-one.

$\left\{\left(1,2\right), \left(2,8\right), \left(3,18\right), \left(4,32\right)\right\}$

Answer to Problem 18AYU

Solution:

Yes, the given function is one–to–one function.

Explanation of Solution

Given:

$\left\{\left(1,2\right), \left(2,8\right), \left(3,18\right), \left(4,32\right)\right\}$

Calculation:

A function is one-to-one if any two different inputs in the domain correspond to two different outputs in the range. That is, if $x_1$ and $x_2$ are two different inputs of a function $f$, then $f$ is one-to-one if $f(x_1)\ne f(x_2)$.

In other way, a function $f$ is one-to-one if no $y$ in the range is the image of more than one $x$ in the domain. A function is not one-to-one if any two (or more) different elements in the domain correspond to the same element in the range.

From the diagram, we can observe that we have 4 different inputs in the domain and also in the range, 4 different outputs, if we map the function, we should get different values as given in the data.

Therefore, different inputs of the domain have the different outputs of the range. As the consequence, the given function is one-to-one.

Therefore $f$ is one-to-one and as if $f(x_1)\ne f(x_2)$.

Also, the function can be as $f\left(x\right)=2x^2$.