#HIII guys.. please help..
24 messages · Page 1 of 1 (latest)
Relations and functions
Dividing side
Ia lready get everything but not this one
Lowk
i believe this is just dividing the function statements directly
The functions are given already given so just divide the given functions accordingly as given and find the domain
Tip: Take the denominator ≠ 0 then you can find the domains easily
According to the example i will show you, solve the problems you have sent.
imagine we have these two functions and we want to find (f/g)(x)
First of all lets take a look at what exactly (f/g)(x) means. It’s basically division, it states that “f(x) is divided by g(x)” or lets say “the stuff in the numerator is divided by the stuff in the denominator”
So as we just said, its the simple division, it means that f(x) is divided by g(x), annnd what are f(x) and g(x)? f(x) is x+1 and g(x) is x-2, therefore we divide these two things. And thats it! We got the function (f/g)(x)
But how about the domain? As we know, domain means the possible x-values we can give as an input, well when we try different numbers in the function (f/g)(x), we see that basically every x-value works! 😱
except one of them… can you guess which one that is?
the number that is NOT in the domain makes the function “undefined”, obviously the numerator can not make the the function undefined(what can it do? Its just there changing its values based on the given x), but the denominator is different, if the denominator becomes zero,the division wouldnt be true anymore!(why?) therefore one of the x-values for this function would NOT let the function be defined and that number is ||2||
basically this means that (f/g)(2) does not exist
the domain is D= R - {2}
(R: Real numbers)
Oh damn the numerator should have been 3, mb
dropping by to tell you you have nice handwriting on a PC/tablet!
thank you xD
The owner can do that 😭