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.

Question

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.

zeesbrat3 · Answer

@Hero @nincompoop @abb0t @kropot72 @Whitemonsterbunny17 @freckles @Miracrown @vera_ewing

zeesbrat3 · Answer

@ganeshie8

zeesbrat3 · Answer

hey

ganeshie8 · Answer

$$f(x)=\dfrac{1-x}{1+x}$$
$$g(x)=\dfrac{x}{1-x}$$

like this ?

zeesbrat3 · Answer

Yes

zeesbrat3 · Answer

$$\frac{ 1- \frac{ x }{ 1-x } }{ 1 + \frac{ x }{ 1-x } }$$

ganeshie8 · Answer

looks good, multply $1-x$ top and bottom

ganeshie8 · Answer

$$\frac{ 1- \frac{ x }{ 1-x } }{ 1 + \frac{ x }{ 1-x } }=
\frac{ \left(1- \frac{ x }{ 1-x }\right)\left(1-x\right) }{ \left(1 + \frac{ x }{ 1-x }\right)\left(1-x\right) } = \dfrac{1-x-x}{1-x+x}=1-2x$$

zeesbrat3 · Answer

why does it seem so much harder?

ganeshie8 · Answer

it is not hard if you simply follow the rules

zeesbrat3 · Answer

i see what i did wrong though

zeesbrat3 · Answer

i multiplied the entire fraction off the bottom, not just the denominator

ganeshie8 · Answer

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 · Answer

Thank you

zeesbrat3 · Answer

Can you help with a few more?
@ganeshie8

zeesbrat3 · Answer

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.