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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

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:

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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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. –  Jack M Apr 4 '13 at 23:16

3 Answers 3

up vote 0 down vote accepted

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? –  Adegoke A Oct 31 '12 at 11:52
    
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. –  MSEoris Oct 31 '12 at 11:55
    
OK. Cool. I also conclude that it can't be surjective because the highest number for the function is 4. Is that right? –  Adegoke A Oct 31 '12 at 11:58
1  
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. –  MSEoris Oct 31 '12 at 12:03
    
Oh I see. Thank you. –  Adegoke A Oct 31 '12 at 12:06

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 $\,8<x+6\,\,,\,\text{ for}\,\,x\in(2,4]\,$ , 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 $\;1-1\;$

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