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reynad

  • one year ago

given f(x)=sinx and g(x)=x^2+1, what is f(g(x)) and g(f(x))?

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  1. jim_thompson5910
    • one year ago
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    \[\Large f(x) = \sin(x)\] \[\Large f(\color{red}{x}) = \sin(\color{red}{x})\] \[\Large f(\color{red}{g(x)}) = \sin(\color{red}{g(x)})\] Do you see how I replaced every x with g(x)?

  2. reynad
    • one year ago
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    that means you would end up with f(g(x))=sin^2x + 1 and using that, g(f(x)) = sin^2x+1 right?

  3. jim_thompson5910
    • one year ago
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    well we replace the g(x) on the right side with x^2 + 1

  4. jim_thompson5910
    • one year ago
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    so let me write out the full steps \[\Large f(x) = \sin(x)\] \[\Large f(\color{red}{x}) = \sin(\color{red}{x})\] \[\Large f(\color{red}{g(x)}) = \sin(\color{red}{g(x)})\] \[\Large f(\color{red}{g(x)}) = \sin\left(\color{red}{x^2+1}\right)\]

  5. jim_thompson5910
    • one year ago
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    As for the other way around, it looks like you got it \[\Large g(x) = x^2 + 1\] \[\Large g(f(x)) = (f(x))^2 + 1\] \[\Large g(f(x)) = (\sin(x))^2 + 1\] \[\Large g(f(x)) = \sin^2(x) + 1\]

  6. jim_thompson5910
    • one year ago
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    no you cannot distribute like that

  7. jim_thompson5910
    • one year ago
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    `sin(x+1)` does NOT turn into `sin(x) + sin(1)`

  8. reynad
    • one year ago
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    wait, so, sin(x^2 +1) cannot be simplified anymore?

  9. jim_thompson5910
    • one year ago
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    yeah it's as simplified as it gets

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