# Confused about one to one functions (concept)

I am confused about the concept of one to one functions when applying it to problems. I know that a one to one function has one x for every y. I was given the function

g(x) = (5x^2 + 15)^7777

and I had to explain why g is not a one to one function. I couldn't really explain the answer but the only thing that came to my mind was because it has a power of 7777. The possible answers are

a) It is not a one to one function because some Y values have two X values.

b) It is not a one to one function because some Y values have three X values.

c) It is not a one to one function because some X values have no Y value.

d) It is not a one to one function because some X values have two Y values.

e) It is not a one to one function because some Y values have one X value.

The answer I chose was d, but it was incorrect. The correct answer was a. I am really confused as to how to apply the concepts to solve this problem. Could someone shed some light on this topic? I would really appreciate it! Thank you.

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If $g(x)$ is a (real) function then a given $x$ defines a unique $y$ by $g(x)$, what you answerd said that there is an $x$ s.t that $g(x)$ can take two values - but that is clearly not the case.

The meaning of $1-1$ is that if $g(x_1)=g(x_2)$ then $x_1=x_2$ - that is it can't be that more then one $x$ is mapped to the same $y$.

Answe $a)$ say that the definition does not hold since there is a $y$ value with $x_1,x_2$ s.t $g(x_1)=g(x_2)$ but $x_1\neq x_2$ . Can you find such example to prove that the function is not $1-1$ ?

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I see my mistake since I misread the answer d. However I am still confused as to how an individual would find how many answers that function has. – Kot Sep 17 '12 at 23:02
What exactly do you mean by "how many answers that function has" ? – Belgi Sep 17 '12 at 23:03
I wanted to know if there was a way to figure out how many x or y values the function had. For example x^2 has two answers. I am a little confused as to how many answers x^7777 has. Is it also two? Is there ever a case where there are three possible answers to a function? – Kot Sep 17 '12 at 23:07
There is no general thing to say here - given a function $g(x)$ and a $y$ value you need to check how many $x$ values are mapped to that $y$, this number can be any number. for example define $g$ by $g(0)=g(1)=g(2):=1$ and for every other number define $g$ to be $0$ – Belgi Sep 17 '12 at 23:11

$\textbf{Hint}$ : $g$ is an even function.

Move cursor over the box for more details.

Note that for all $x$, $g(x) = g(-x)$. This is because $g(x) = (5x^2 + 15)^{7777} = (5(-x)^2 + 15)^{7777} = g(-x)$. Hence, you have, for example, $g(1) = g(-1)$. So $f$ is not injective because both $1$ and $-1$ map to the same thing.

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I know that a one to one function has one x for every y.

It is at most one x for every y. Then you can see why the above is not one to one.

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