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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 http://gyazo.com/d5e86e8598726d2c5be2a28578deaeeb.png?1351682833

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:

http://gyazo.com/f6441e550637aa9fe15a40779405d989.png?1351682990

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.

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    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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    To get injective, I got the elements, 0,1,8, 9, & 10. These number are of the second set, so therefore it is injective. Right?2012-10-31
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    looks good to me, though keep in mind, you are technically supposed to be checking every single value in the ranges (not just whole numbers), so the perspective offered by Peter or DonAntonio are more technically sound.2012-10-31
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    OK. Cool. I also conclude that it can't be surjective because the highest number for the function is 4. Is that right?2012-10-31
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    not quite, the function is indeed surjective because for x values between $[0,2]$, $f(x)$ goes up to 8, and $[2,4]$, $f(x)$ goes from a little more than 8 to 10, thus the whole range [0,10] is covered by the function.2012-10-31
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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 $\;1-1\;$