x ≠ -2 is true for f(x), but we are interested in f[g(x)]. In that case, the restriction is _g(x)_ ≠ -2. Solve that to find the restriction in terms of x. 4/(x - 1) ≠ -2 -2(x - 1) ≠ 4 x - 1 ≠ -2 x ≠ -1 This is the same restriction that already applies to f[g(x)] by its own definition, so an additional note is not necessary.
I have some worry if we consider fog as a one function h definition Domaine is R-{-1} h(x) = (x-1)/2x+2 but f(g(x)) =f(4/x-1) is defined when g is defined and f is defined. But fog or h is defined because x-1 come to the top of fraction.
honestly is a little bit confusin I asked him around it. the combination have an image and g is undefined in the point and f is continue and have a limite in infinity.
I love your videos because they are special, not obvious. The information you bring is ALWAYS interesting. Great video
I like your contant is clear and methodical.
Thank you sir!! I can use this on my upcoming licensure exam💗🙏🏻
I'm very surprised to learn something like this 😮❤❤
Excellent 👌
Nice question sir
Nice one
Thank you sir
Sir make a video for finding domain and range in a questions containing:-
log and trigonometric functions
Well asked
Why do you not include the restriction of x != -2 from the f(x) function?
I think because the question is asking specifically for f(g(x))
x ≠ -2 is true for f(x), but we are interested in f[g(x)]. In that case, the restriction is _g(x)_ ≠ -2. Solve that to find the restriction in terms of x.
4/(x - 1) ≠ -2
-2(x - 1) ≠ 4
x - 1 ≠ -2
x ≠ -1
This is the same restriction that already applies to f[g(x)] by its own definition, so an additional note is not necessary.
I have some worry if we consider fog as a one function h definition Domaine is R-{-1} h(x) = (x-1)/2x+2
but f(g(x)) =f(4/x-1) is defined when g is defined and f is defined.
But fog or h is defined because x-1 come to the top of fraction.
My question is lim(x->(+or-)infinity) f(x) exist and f is continue. can we tolerate f(g(x)) when g-> (+or-) infinity?
❤❤❤
When x=1, f(g(x))=0 so why exclude x=1 from the domain?
Year 9 student here but I'm pretty sure it's because if x=1, then the denominator of g(x) would be 0
I wondered the same thing. The excluded values should be x cannot equal {-2, 1}, right? They create vertical asymptotes.
honestly is a little bit confusin I asked him around it. the combination have an image and g is undefined in the point and f is continue and have a limite in infinity.
@@malforon4893 right
@@gregnixon1296 -2 and 1 are the excluded values of f(x) and g(x) but not f(g(x)) where the excluded value is -1.
Please more limit videos aaa
Holy smokes 2 got 1 right how did that happen
f(x) ; {x|x-2 }?