Shayaan_Mustafa
  • Shayaan_Mustafa
f(x)=2sqrt(x-1) and g(x)=sqrt(x-1) Find and state domain f(x)*g(x)
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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Shayaan_Mustafa
  • Shayaan_Mustafa
i have answer i.e. 2x-2 but what about domain?
apoorvk
  • apoorvk
domain is what values x may take in a function for being defined. here x can take any real value for being defined no constraints so domain is all real nos.
Shayaan_Mustafa
  • Shayaan_Mustafa
yes i write as you said. but the answer is wrong. not all x's

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amistre64
  • amistre64
sqrt(n); n>=0
Shayaan_Mustafa
  • Shayaan_Mustafa
ok amistre64 how can you say? but look at answer. there is not any sqrt in answer. simple 2x-2
amistre64
  • amistre64
the domain of the function dont change since they are in integral part of the problem so we have to "and" theyre respective domains together
Shayaan_Mustafa
  • Shayaan_Mustafa
it is a questio from howard anton calculus. and according to book domain should be x>=1
apoorvk
  • apoorvk
but in f(x), |dw:1332517010644:dw| because anything under a sq root can not be negative. so x has to be greater than one which is the domain of f(x). similar in g(x). oh yes.. hmm in f*g, although the final function is 2x-2 but the original factors are in roots right? so x will have to be greater than 1 there as well. but if someone asks you simply whats domain 2x-2 then we can say all real.
amistre64
  • amistre64
f(x)=2sqrt(x-1) ; D = {x|x >= 1} g(x)=sqrt(x-1) ; D = {x|x >=1}
amistre64
  • amistre64
-2-1 is NOT >= 0
Shayaan_Mustafa
  • Shayaan_Mustafa
no no wait you guys. try to make me understand.
amistre64
  • amistre64
so yes, the intersection of both domains is ... well {x|x >= 1} :)
amistre64
  • amistre64
Do you see that the domain of f and g are the same?
amistre64
  • amistre64
the domain of sqrt(x-1) is the same as the domain of sqrt(x-1) ... so all we have to do is determine the domain of sqrt(x-1)
Shayaan_Mustafa
  • Shayaan_Mustafa
look. f(x) is a function which is under sqrt i.e. 2sqrt(x-1) and g(x) as well i.e. sqrt(x-1) and we have said to find f(x)*g(x) i.e. the multiple of both functions not separately. So therefore we will state the domain of f(x)*g(x)=2x-2. and here is not anything under sqrt? What is this. this is irritating me.
Shayaan_Mustafa
  • Shayaan_Mustafa
no no. look at quest. we have to find domain of f(x)*g(x) not separate functions
amistre64
  • amistre64
if I tell you to put a rat and a cat into the same box; does that convert them into a ratcat hybrid?
amistre64
  • amistre64
your separate function are what controls your NEW f*g function regardless of what it simplifies to
Shayaan_Mustafa
  • Shayaan_Mustafa
hahahaha... ofcourse not. lol so it is that you are trying to say??
Shayaan_Mustafa
  • Shayaan_Mustafa
yes. that's the thing for what i am asking for. thanks a lot.
amistre64
  • amistre64
x^2/x still goes bad at x=0 even tho it simplifies to x
Shayaan_Mustafa
  • Shayaan_Mustafa
ok its mean whenever i say to find of a composite functions either then I always look for original function's domain. Right?
amistre64
  • amistre64
yes, and the composite is the "common" points of the domains of your separate functions
amistre64
  • amistre64
it just so happens in this one that they have the exact same domains :)
Shayaan_Mustafa
  • Shayaan_Mustafa
yes. i got. thnx again.
amistre64
  • amistre64
and, if you have to do a division of functions; then your new function has to look at the zeros of the denominator as well ....
Shayaan_Mustafa
  • Shayaan_Mustafa
yes yes. i know.

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