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I'm trying to do my Maths assignment, I looked at the lecturer's notes for examples but it seems like at lot of steps at skipped. Are is the example:

pic
(source: gyazo.com)

I understand what Injective and Surjective functions are by watching this video:

http://www.youtube.com/watch?v=xKNX8BUWR0g

But I'm still not able to do my assignment, which is this:


(source: gyazo.com)

So far I have this:

For $f[0,4], x^3:$

$f(0) = 0^3 = 0$

$f(1) = 1^3 = 1$

$f(2) = 2^3 = 8$

$f(4) = 4^3 = 64$

For $f[0,4], x + 6:$

$f(0) = 0 + 6 = 6$

$f(1) = 1 + 6 = 7$

$f(2) = 2 + 6 = 8$

$f(3) = 3 + 6 = 9$

$f(4) = 4 + 6 = 10$

I'm not sure what do to next.

Thank you.

  • 1
    Draw a graph. Does the function hit every y-coordinate in [0, 10]? If (and only if) so, it's surjective. Do any two x-coordinates map to the same y-coordinate? If (and only if) so, it's not injective. Once you can see it on the graph, it should be easy to work out a proof.2013-04-04

3 Answers 3

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To check injective : every element in your domain needs to correspond to a unique element in your range.

Thus you must check that $f(1) \ne f(2) \ne f(3) \ne f(4)$.

Note that for $f(3),f(4)$ you need to use the function $x+6$, while for $f(1), f(2)$ you must use $x^3$ (take care with the less than or equal to symbol)

To check surjective, you need to make sure that every element in $[0,10]$ can be attained through one of your functions defined on $[0,4]$

If you have both, then your function is bijective.

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    it's not just that the answers of Peter and DonAntonio are "more technically sound". Checking that $f(1)\neq f(2)\neq f(3)\neq f(4)$ is no justification at all for the claim that $f$ is injective. This check would be sufficient if the domain of $f$ were the set $\{1,2,3,4\}$ instead of the interval $[0,4]$.2013-01-04
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Imagine its graph. $f$ is increasing on $[0,2]$, mapping it surjectively to $[0,8]$. Then on $(2,4]$, it is again increasing, mapping surjectively to $(8,10]$. Putting these together, you can see $f:[0,4]\rightarrow[0,10]$ is surjectively. That is, it "hits" every value in $[0,10]$. Injectivity is a bit harder, but you can see that $f$ is a strictly increasing function on $[0,4]$, so it's injective.

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First: Clearly $\,0\leq x^3\leq 8\,\,,\,\text{ for}\,\,x\in [0,2]\,$, and also $\,8 , so that both branches of definition of $\,f\,$ don't "mingle".

Second: For $\,x,y\in[0,2]\,$ , we get:

$x^3=y^3\Longleftrightarrow (x-y)(x^2+xy+y^2)=0\Longleftrightarrow x=y$

since the large expression above is clearly positive or zero, so the part of $\,f\,$ which equals the cubic is $\;1-1\;$, and since clearly the second part is an ascending line, the whole function is $\;1-1\;$