Please help:)
So I had to simplify \[\frac{ 4x ^{2}+7-2 }{ 4x-1 }\]that part is done but now it's asking explain why f(1) = 3, f(0) = 2, and f(-1) = 1, yet f\[\frac{ 1 }{ 4 }\] is undefined.
well, if the denominator of a fraction is zero, then the number is undefined
just looking at the denominator 4x-1 what happens if x=1/4?
and i also assume its possible to cancel out the denominator, but that only works if what you're canceling out is equal to 1, not zero
so it would be like this? 4 (1/4) -1=1?
long story short 1/1= 1 0/0 is not 1
... 4 times 1/4 is?
in other words, 4 divided by 4 is?
1
ok now 1-1 is?
0 which the reason it is undefined correct?
ok now, lets look at trends 1/1= 1 1/.1 =10 1/.01 =100 1/.001=1000 so lets assume 0 is equivalent to a really really low number, like .0000000000000000000000001 so what happens when we get 1/0 ?
Undefined?
well, we get an infinitely large number or as some would call it, infinity but infinity is not an actual number, thus which is why we considered it undefined
so basically, if you ever get a fraction whose denominator is zero, then the solution is undefined however, if you're dealing with limits, then its infinity a third note 0/0 is not equal to 1
ok back to your previous question, do you think you can answer it now?
Maybe only for the reason why 1/4 is undefined but not the others. I don't understand why it's asking to explain why f(1) = 3, f(0) = 2, and f(-1) = 1
ok, so basically, what you were asked to simplify, is what is being considered as a function f(x) so when x =1, then the fraction is equal to 3 when x-0, then the fraction is equal to 2 your question is asking, why do those values have actual answers, when x=1/4 does not
So pretty much it's like saying 4(1)-1=3 I don't get how f(0)=2 and f(-1)=1
no, the function is the entire fraction, not just the denominator
f(0)= (4*0^2-7(0)-2)/(4(0)-1)=?
2 OH OKAY
I get it now:)Thank you very much!!
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