Maximal Domain Convention

Apr 2010
59
3
Hey its me again;

Im going through my lecture notes so I can get a headstart on next years work, 1 moedule in particular has been very poorly organized and there is no working for a lot of questions, just the question written as example and I know for a fact we did not go over them in class.

Iv contacted our module leader and had no reply (2 days) so im getting kinda anxious, so heres the question, it was marked as Example 4; but with no working or no notes attached.

 
Oct 2009
4,261
1,836
Hey its me again;

Im going through my lecture notes so I can get a headstart on next years work, 1 moedule in particular has been very poorly organized and there is no working for a lot of questions, just the question written as example and I know for a fact we did not go over them in class.

Iv contacted our module leader and had no reply (2 days) so im getting kinda anxious, so heres the question, it was marked as Example 4; but with no working or no notes attached.



This question belongs to precalculus!(Punch)(Angry)

You have to check whether \(\displaystyle f(x)=f(y)\Longrightarrow x=y\) , and you can do this by two ways:

The long way: \(\displaystyle f(x) =\frac{x^2-3x+2}{x^2+3x+2}=\frac{y^2-3y+2}{y^2+3y+2}=f(y)\) . Cross multiply this, do a little algebra and reach, finally, \(\displaystyle f(x)=f(y)\Longrightarrow (xy-2)(x-y)=0\) , and this must suffice to get the answer.

The short way: factor both the numerator and the denominator of the function: \(\displaystyle f(x)=\frac{x^2-3x+2}{x^2+3x+2}=\frac{(x-1)(x-2)}{(x+1)(x+2)}\) , and you're (almost) done.
Hint: the function isn't 1-1

About onto: you've to check whether \(\displaystyle f(x)=w\iff \frac{x^2-3x+1}{x^2+3x+2}=w\iff (w-1)x^2+3(w-1)x+2(w-1)=0\) has a real solution for some given real number \(\displaystyle w\). This is a simple exercise in quadratic and you have to check for which real values of \(\displaystyle w\) the above quadratic's discriminant is non-negative.
Hint: the function is onto

Tonio
 
Apr 2010
59
3
Erm, reading the example again - what you have done on the "short method" is right, I agree with that, its where I was starting with it.

I have done this atm: \(\displaystyle f(x) = y\) and from that used the fact that there are 4 values that \(\displaystyle x\) could be to give a value for \(\displaystyle f(x)\)

I came to the conclusion that it was Surjective and not Injective or Bijective.

The next part says demonstrate this by using 2 values of x, x1 and x2. From the short method, I use \(\displaystyle x = 1\) and \(\displaystyle x = 2\), I don't use \(\displaystyle x = -1\) and \(\displaystyle x = -2\) because that would give a 0 on the denominator so that would not exist.

By putting: \(\displaystyle x = 1\) and \(\displaystyle x = 2\) into \(\displaystyle f(x)\) I get them both to equal each other at 0.

Not it also says to find the range, but how do I go about that?
 
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