# Thread: Proving a function is onto.

1. ## Re: Proving a function is onto.

Originally Posted by mathproblems
yes, this is what I am saying. I went to school 25 years ago...now I don't remember some things. And now I am taking discrete math ...
Please go back and sit in of a basic algebra class.
If you do not, you are going to be frustrated beyond belief by the material you will meet in a discrete mathematics course.

2. ## Re: Proving a function is onto.

too late, I am already in the middle of semester.

3. ## Re: Proving a function is onto.

Originally Posted by mathproblems
too late, I am already in the middle of semester.
Then maybe you would benefit from some 1-1 help?

(pun intended)

4. ## Re: Proving a function is onto.

yes, I got in a person tutor. I still try to do my homework by myself first.

anyway, could someone please explain the f (n) =?

5. ## Re: Proving a function is onto.

we need to find an x so that f(x) = c, for any c we choose. so x should be "some expression (formula) in c".

what is f(x)? by the definition:

$\displaystyle f(x) = 2x^3 - 4$. so if f(x) = c, we have:

$\displaystyle 2x^3 - 4 = c$. now, we need to try to "solve for x".

$\displaystyle 2x^3 = c+4$ (adding 4 to both sides)

$\displaystyle x^3 = \frac{c+4}{2}$ (dividing both sides by 2)

$\displaystyle x = \sqrt[3]{\frac{c+4}{2}}$ (taking the cube root of both sides). this is the x we are looking for (we hope).

to check, we verify that f(x) is indeed c, when $\displaystyle x = \sqrt[3]{\frac{c+4}{2}}$.

$\displaystyle f\left(\sqrt[3]{\frac{c+4}{2}}\right) = 2\left(\sqrt[3]{\frac{c+4}{2}}\right)^3 - 4$

$\displaystyle = 2\left(\frac{c+4}{2}\right) - 4 = (c+4) - 4 = c + (4-4) = c+0 = c$,

so we indeed found an x that f sends to c.

6. ## Re: Proving a function is onto.

oh wow!! Thank you!!! Now I got it!

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