Shayaan_Mustafa
  • Shayaan_Mustafa
f(x)=2x+1 and g(x)=x^(2)-x Calculate fog and gof and domains of the functions.
Mathematics
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chestercat
  • chestercat
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Shayaan_Mustafa
  • Shayaan_Mustafa
help here.
anonymous
  • anonymous
\[f(x)=2x+1\] \[g(x)=x^2-x\] \[f\circ g(x)=f(g(x))=f(x^2-x)=2(x^2-x)+1\] then maybe algebra
anonymous
  • anonymous
\[f\circ g(x)=2x^2-2x+1\] domain all real numbers because you are working with polynomials

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Shayaan_Mustafa
  • Shayaan_Mustafa
yes ok . I have solved the composite function. problem for domain. how to find domain?
Shayaan_Mustafa
  • Shayaan_Mustafa
ok all real numbers except inf. right?
anonymous
  • anonymous
step one: get rid of circle notation \[g\circ f(x)=g(f(x))\] step two: replace the general f by the specific one you have \[g(f(x))=g(2x+1)\] step three: evaluate the function at your new input, that is replace "x" by whatever is in the parentheses \[g(2x+1)=(2x+1)^2-(2x+1)\] step four: clean up with algebra
anonymous
  • anonymous
all real numbers is all real numbers infinity not a number
Shayaan_Mustafa
  • Shayaan_Mustafa
i know how to solve composite function. someone said me domain is located as individual functions not combined functions. for example. f(x)=2sqrt(x-1) and g(x)=sqrt(x-1) and want to calculate f(x)/g(x) so f(x)/g(x) = 2sqrt(x-1)/sqrt(x-1) so here we will look at denominator for domain. although function is not fully computed. if i complete my calculation then result would be f(x)/g(x)=2 so here we can't estimate domain. But in my given question. First we should solve the composite function and then find domain, not like above example. right?
anonymous
  • anonymous
no you are right, you need the domain of the "inside" function first i ignored that because you are working with polynomials so domain is all real numbers for each, and also for the composite
Shayaan_Mustafa
  • Shayaan_Mustafa
f(g(x))=2(x^2-x)+1 you mean at this stage we have to locate domain. right?
anonymous
  • anonymous
for example if \[f(x)=x^2\] and \[g(x)=\sqrt{x-1}\] then \[f\circ g(x)=f(g(x))=f(\sqrt{x-1})=\sqrt{x-1}^2=x-1\] but the domain is determined by the domain of \[g(x)\] which is \[[1,\infty)\]
Shayaan_Mustafa
  • Shayaan_Mustafa
OK dude. I got you. for composite functions domain will always defined by taking inner function. Thanks dude.

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