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My first step was to isolate the 5 so that 0=(5(1-x^2)/(x^2+1)).
Hi Panavia, An equation of the form (top ) / (bottom) = zero Must obey TWO rules: 1) (bottom) CANNOT equal zero! 2) (top) MUST equal zero. So in this case: 1) (bottom) = x^2 +1 , which is never zero, so we're good for rule one. 2) (top) = 5-5x^2 So the question is, how do we solve: \[5-5x^2=0\] Doing this isn't so bad, just factor out the 5 \[5(1-x^2)=0\] and now solve \[1-x^2 = 0\] This is pretty easy, just add x^2 to both sides \[1=x^2\] and the answer should be right in front of you. :)
Yes, that's what I ment with logical reasoning. But when I plot the graph I get two zeros. One at +1 and one at -1. The -1 is the one which puts me off. Like you said, when I put the -1 into the bottom portion, I get a 0 in the denominator...
just a comment, you could have p(x)/q(x) = 0/0 , in which case you can look to factor out something or analyze it further , look at the limit as x approaches your zeroes
but i wouldnt go so far as to say the bottom cannot equal to zero
Hmm, so use the L'Hospital rule?
no, i was just talking in general
in this case the denominator is strictly positive
he said it must obey two rules. there are exceptions
The reason you get -1 and +1 as your answer is because we can factor: \[1-x^2=(x+1)(x-1)\] Or when we solve \[1=x^2 \Rightarrow x=\pm1\] Substituting -1 into the denominator does not yield zero: \[(-1)^2+1 =1+1 =2\] Remember when you square negative one, or multiply -1 times itself, you get a positive.
for example find 0 = (x^2 + x - 6) /(x-2)
err, solve that i mean
Oh, yes. You are right. I forgott that. Thanks a lot!
That's pretty easy. In this case you can really solve the equation. But in my case it seems like there is no such solution and you can only figure out the zeros by logic reasoning...
But thanks a buch! You guys helped me a lot!
just because you have zero in the denominator does not mean you must stop there and say no solution, like 1/x = 0 (no solution). for my example 0 = (x^2 + x - 6) /(x-2) you have 0/0 for x = 2, but we still have x=-3 as a solution
so my point is, look out for 0/0
Ok, thanks. So it seems like the functions just have gaps at certain points...
yes there could be a hole, or a jump discontinuity
an essential discontinuity or a removable (if i plug it) discontinuity