Answer:
Because its graph will not pass the vertical line test.
Step-by-step explanation:
We must first find the inverse of the given functions.
Let
[tex]y=f(x)[/tex].
Then for the first function, we have [tex]y=x^2[/tex]
We interchange x and y to get,
[tex]x=y^2[/tex]
We make y the subject to get,
[tex]\pm \sqrt{x}=y[/tex]
This is not a function because one x-value is mapping onto two y-values.
Hence its graph will not pass the vertical line test.
See red graph.
For the second function, we again let
[tex]y=g(x)[/tex]
Then,
[tex]y=x^3[/tex]
We interchange x and y to get,
[tex]x=y^3[/tex]
We make y the subject to get,
[tex]x^{\frac{1}{3}}=y[/tex]
This is a function because, one x-value maps on to one and only one y-value. This tells us that the graph of this function will past the vertical line test.
see blue graph.
