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zeesbrat3

  • one year ago

For f of x equals the quotient of the quantity 1 minus x and the quantity 1 plus x and g of x equals the quotient of the quantity x and the quantity 1 minus x, find the simplified form for f [g(x)] and state the domain.

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  1. zeesbrat3
    • one year ago
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    @Hero @nincompoop @abb0t @kropot72 @Whitemonsterbunny17 @freckles @Miracrown @vera_ewing

  2. zeesbrat3
    • one year ago
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    @ganeshie8

  3. zeesbrat3
    • one year ago
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    hey

  4. ganeshie8
    • one year ago
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    \[f(x)=\dfrac{1-x}{1+x}\] \[g(x)=\dfrac{x}{1-x}\] like this ?

  5. zeesbrat3
    • one year ago
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    Yes

  6. zeesbrat3
    • one year ago
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    \[\frac{ 1- \frac{ x }{ 1-x } }{ 1 + \frac{ x }{ 1-x } }\]

  7. ganeshie8
    • one year ago
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    looks good, multply \(1-x\) top and bottom

  8. ganeshie8
    • one year ago
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    \[\frac{ 1- \frac{ x }{ 1-x } }{ 1 + \frac{ x }{ 1-x } }= \frac{ \left(1- \frac{ x }{ 1-x }\right)\left(1-x\right) }{ \left(1 + \frac{ x }{ 1-x }\right)\left(1-x\right) } = \dfrac{1-x-x}{1-x+x}=1-2x\]

  9. zeesbrat3
    • one year ago
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    why does it seem so much harder?

  10. ganeshie8
    • one year ago
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    it is not hard if you simply follow the rules

  11. zeesbrat3
    • one year ago
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    i see what i did wrong though

  12. zeesbrat3
    • one year ago
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    i multiplied the entire fraction off the bottom, not just the denominator

  13. ganeshie8
    • one year ago
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    For domain part, notice that \(g(x)=\dfrac{x}{1-x}\) is undefined at \(x=1\) and since \(1-2x\) is defined for all real numbers, the domain of \(f(g(x))=1-2x\) is all real number except \(1\)

  14. zeesbrat3
    • one year ago
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    Thank you

  15. zeesbrat3
    • one year ago
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    Can you help with a few more? @ganeshie8

  16. zeesbrat3
    • one year ago
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    A particle moves on a line away from its initial position so that after t hours it is s = 2t2 +3t miles from its initial position. Find the average velocity of the particle over the interval [1, 4]. Include units in your answer.

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