Question

In question 1, you found the cubes of both positive and negative numbers. Does x = 8 have two solutions as x2 = 4 does? Why or why not?

Answer 1

No.

1) Because in the cubic root of 8, as below represented in this function:

`\begin{gathered} f(x)\text{ =}\sqrt[3]{x} \\ f(8)\text{ =}\sqrt[3]{8}\text{ = 2} \\ f(-8)\text{ =}\sqrt[3]{-8\text{ }}\text{ =-2} \end{gathered}`

There is no other negative number that can be inserted in the Domain to yield a positive value in the Range.

Unlike, the quadratic radical function. For example:

`\begin{gathered} f(x)\text{ =}√(x) \\ f(25)\text{ =+5 or -5} \\ (-5)^2=25and(5)^2=25 \end{gathered}`