zeesbrat3
  • zeesbrat3
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zeesbrat3
  • zeesbrat3
@Hero @nincompoop @abb0t @kropot72 @Whitemonsterbunny17 @freckles @Miracrown @vera_ewing
zeesbrat3
  • zeesbrat3
@ganeshie8
zeesbrat3
  • zeesbrat3
hey

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More answers

ganeshie8
  • ganeshie8
\[f(x)=\dfrac{1-x}{1+x}\] \[g(x)=\dfrac{x}{1-x}\] like this ?
zeesbrat3
  • zeesbrat3
Yes
zeesbrat3
  • zeesbrat3
\[\frac{ 1- \frac{ x }{ 1-x } }{ 1 + \frac{ x }{ 1-x } }\]
ganeshie8
  • ganeshie8
looks good, multply \(1-x\) top and bottom
ganeshie8
  • ganeshie8
\[\frac{ 1- \frac{ x }{ 1-x } }{ 1 + \frac{ x }{ 1-x } }= \frac{ \left(1- \frac{ x }{ 1-x }\right)\left(1-x\right) }{ \left(1 + \frac{ x }{ 1-x }\right)\left(1-x\right) } = \dfrac{1-x-x}{1-x+x}=1-2x\]
zeesbrat3
  • zeesbrat3
why does it seem so much harder?
ganeshie8
  • ganeshie8
it is not hard if you simply follow the rules
zeesbrat3
  • zeesbrat3
i see what i did wrong though
zeesbrat3
  • zeesbrat3
i multiplied the entire fraction off the bottom, not just the denominator
ganeshie8
  • ganeshie8
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 \(1\)
zeesbrat3
  • zeesbrat3
Thank you
zeesbrat3
  • zeesbrat3
Can you help with a few more? @ganeshie8
zeesbrat3
  • zeesbrat3
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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