reynad
  • reynad
given f(x)=sinx and g(x)=x^2+1, what is f(g(x)) and g(f(x))?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
\[\Large f(x) = \sin(x)\] \[\Large f(\color{red}{x}) = \sin(\color{red}{x})\] \[\Large f(\color{red}{g(x)}) = \sin(\color{red}{g(x)})\] Do you see how I replaced every x with g(x)?
reynad
  • reynad
that means you would end up with f(g(x))=sin^2x + 1 and using that, g(f(x)) = sin^2x+1 right?
jim_thompson5910
  • jim_thompson5910
well we replace the g(x) on the right side with x^2 + 1

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jim_thompson5910
  • jim_thompson5910
so let me write out the full steps \[\Large f(x) = \sin(x)\] \[\Large f(\color{red}{x}) = \sin(\color{red}{x})\] \[\Large f(\color{red}{g(x)}) = \sin(\color{red}{g(x)})\] \[\Large f(\color{red}{g(x)}) = \sin\left(\color{red}{x^2+1}\right)\]
jim_thompson5910
  • jim_thompson5910
As for the other way around, it looks like you got it \[\Large g(x) = x^2 + 1\] \[\Large g(f(x)) = (f(x))^2 + 1\] \[\Large g(f(x)) = (\sin(x))^2 + 1\] \[\Large g(f(x)) = \sin^2(x) + 1\]
jim_thompson5910
  • jim_thompson5910
no you cannot distribute like that
jim_thompson5910
  • jim_thompson5910
`sin(x+1)` does NOT turn into `sin(x) + sin(1)`
reynad
  • reynad
wait, so, sin(x^2 +1) cannot be simplified anymore?
jim_thompson5910
  • jim_thompson5910
yeah it's as simplified as it gets

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