# Thread: Help with 1-1, onto fuctions; and functions! (emergency)

1. ## Help with 1-1, onto fuctions; and functions! (emergency)

1) Let F be the function from R to R defined by f(x)=x^2. Find:
a) f^(-1) ({1})
b) f^(-1) ({x | 0 < x < 1})
c) f^(-1) ({x | x > 4})

2) Give an example of a function from N to N that is
a) one-to-one but not onto.
b) onto but not one-to-one.
c) both onto and one-to-one (but different from the identity function).
d) neither one-to-one nor onto.

Thank You very much in advance.
Anu

2. Originally Posted by anu
1) Let F be the function from R to R defined by f(x)=x^2. Find:
a) f^(-1) ({1})
b) f^(-1) ({x | 0 < x < 1})
c) f^(-1) ({x | x > 4})
Okay, so the equation is $f(x)=x^{2}$

a) $f^{-1}(\{1\})$ is saying what value of f(x) will result in an answer where the inverse has a domain of 1. This sounds more complex than it is, because the range of the function is the domain of the inverse, all they're saying is what value of x will give you a y value of 1. Obviously there are two values for x that when squared give you 1. They would be positive and negative 1.

b) $f^{-1}(\{x | 0 < x < 1\})$ Okay, so just reword this to be what x values for f(x) will give you y values greater than 0 and less than 1. Well $0^{2} = 0$ and $1^{2}=1$ So you know that x values between 0 and 1 will give you y values between 0 and 1. Now, because the sign is greater than, but not equal to, and less than but not equal to, you know that x can never actually be 0 or 1. So the answer would be (0, 1) in parentheses.

c) $f^{-1}(\{x | x > 4\})$ Right, so what x values when plugged into the equation $f(x)=x^{2}$ will give you y values that are greater than 4?

3. Originally Posted by anu
1) Let F be the function from R to R defined by f(x)=x^2. Find:
a) f^(-1) ({1})
to find $f^{-1}(x)$ for x in the range of a function, is to find the element in the domain that maps to it. that is, if $f^{-1}(y)$ exists and $f^{-1}(y) = x$ then $f(x) = y$

so here, we want the element in the domain that maps to the element 1. for $f(x) = x^2$, there are two elements that map to one. thus: $f^{-1}( \{1 \}) = \{ 1, -1 \}$

since $f(1) = f(-1) = 1$

b) f^(-1) ({x | 0 < x < 1})
here we want all the elements that map to the set {x | 0 < x < 1}. these are the values in the interval (-1,1) in the domain. do you see why?

so $f^{-1}( \{ x | 0 < x < 1 \}) = \left \{ y | -1 < y < 1, y \in \mathbb{R} \right \}$

c) f^(-1) ({x | x > 4})
try this one by yourself.

2) Give an example of a function from N to N that is
a) one-to-one but not onto.
take the even integers that are in $\mathbb{N}$.

so $f(n) = 2n$ for $n \in \mathbb{N}$ works. can you come up with any more functions? why is this function 1-1 but not onto?

b) onto but not one-to-one.
how about $f(n) = \left \{ \begin{array}{lr} 1 , & \mbox{ if } n = 1 \\ n - 1, & \mbox { if } n \ge 2 \end{array} \right.$ for $n \in \mathbb{N} = 1,2,3,...$

why is this function onto but not 1-1?

c) both onto and one-to-one (but different from the identity function).
$f(n) = \left \{ \begin{array}{lr} 2 , & \mbox{ if } n = 1 \\ 1, & \mbox { if } n = 2 \\ n, & \mbox { if } n \ge 3 \end{array} \right.$ for $n \in \mathbb{N} = 1,2,3,...$

d) neither one-to-one nor onto.
you try this one

EDIT: Too slow, yet again

4. Originally Posted by angel.white
c) $f^{-1}(\{x | x > 4\})$ Right, so what x values when plugged into the equation $f(x)=x^{2}$ will give you y values that are greater than 4?
x > 2, right?

5. Originally Posted by anu
x > 2, right?
what about, say, x = -5?

6. d) neither one-to-one nor onto.

f(n) = 1
then the function is neither 1-1 nor onto, right?

Originally Posted by Jhevon
what about, say, x = -5?
Oh ya so x > 2 and x < -2

7. Righto ^_^

Of course, you need to put it in proper notation for your instructor, but I'm guessing you can handle that :P

And I'm impressed with how elegantly Jhevon answered the second question, I was sitting here laboring over it, only just finished question b when I realized it'd already been answered.

edit: I'm guessing you're using the book Discreet Mathematics and It's Applications by Kenneth H. Rosen? I'm actually taking that class right now myself this was in section 2.3, which will be on my test coming up this Wednesday

8. Originally Posted by anu
d) neither one-to-one nor onto.

f(n) = 1
then the function is neither 1-1 nor onto, right?
correct

Oh ya so x > 2 and x < -2
again, correct. but it seems your questions demand notation that you are not employing here. use set notation to give the answer