Could someone please explain to me how you can tell if a function is one-to-one? I don't understand...

2. Originally Posted by runner07
Could someone please explain to me how you can tell if a function is one-to-one? I don't understand...
do you know what it means to be one-to-one? what is you're definition of one-to-one functions.

graphically: we can show that a graph is a one-to-one relation by the horizontal line test. drawing any horizontal line will cut the graph ONLY once

analytically: we can show a graph $\displaystyle f(x)$ is one-to-one by showing that if $\displaystyle f(x_1) = f(x_2)$ then $\displaystyle x_1 = x_2$

or equivalently, if we can find any two none equal elements in the domain, $\displaystyle x_1 \neq x_2$ then we would find that $\displaystyle f(x_1) \neq f(x_2)$

3. i understand that one-to-one is using the horizontal line test but i cant always draw a graph to check. its the analytical part that confuses me.

4. Originally Posted by runner07
i understand that one-to-one is using the horizontal line test but i cant always draw a graph to check. its the analytical part that confuses me.
example:

the function y = x + 1 is one-to-one, since if we have:

$\displaystyle x_1 + 1 = x_2 + 1$ (which is $\displaystyle f(x_1) = f(x_2)$)

we can subtract 1 from both sides to get:

$\displaystyle x_1 + 1 - 1 = x_2 + 1 - 1$

$\displaystyle \Rightarrow x_1 = x_2$

thus the function is one-to-one

counter-example:

the function $\displaystyle y = x^2$ is not one-to-one

since if we have $\displaystyle f(x_1) = f(x_2)$ it is not necessarily true that $\displaystyle x_1 = x_2$

look: $\displaystyle x_1^2 = x_2^2$
square rooting both sides we obtain:

$\displaystyle x_1 = \pm x_2$ in general

a specific example would be, $\displaystyle f(1) = f(-1)$ since $\displaystyle 1^2 = (-1)^2 = 1$ (you could just state this example by the way to prove it's not one-to-one, no explanation needed)

so there are two elements in the domain that map to a single element in the range. thus we have a many-to-one relation, not one-to-one

5. when you take the y=x^2 and you do this..

x1^2 = x2^2

why is it only the x2 that becomes +- when you square root it? why doesnt x1 become +- too?

6. Originally Posted by runner07
when you take the y=x^2 and you do this..

x1^2 = x2^2

why is it only the x2 that becomes +- when you square root it? why doesnt x1 become +- too?
that's why i said "in general."

let's say i did do that:

$\displaystyle \pm x_1 = \pm x_2$

how many possible combination of signs can we have? well 4 of course

1. $\displaystyle x_1 = x_2$

2. $\displaystyle x_1 = - x_2$

3. $\displaystyle -x_1 = -x_2$

4. $\displaystyle -x_1 = x_2$

Of course, we realize that 1 and 3 are saying the same thing, and 2 and 4 are saying the same thing. so i can summarize these 4 possibilities into two possibilities

1. $\displaystyle x_1 = x_2$ .............(this is equation 1 above, which is the same as equation 3)

2. $\displaystyle x_1 = -x_2$ ...........(this is equation 2 above, which is the same as equation 4)

Now I can summarize these two equations into one:

namely, $\displaystyle x_1 = \pm x_2$

but like i said, this kind of reasoning is not necessary, saying "since $\displaystyle f(1) = f(-1)$, we have that $\displaystyle f(x)$ is NOT one-to-one" will suffice