Question 109845
Let f(x)=4-x^ ….do you have here {{{x^2}}} or {{{x}}}?


if you have {{{x^2}}}, then we have:

{{{f(x) = 4-x^2}}}
{{{g(x) = 2-x}}}

Then: {{{f/g(x) = (4-x^2)/( 2-x)}}}……….in nominator, {{{4}}} we can write as {{{2^2}}}

{{{f/g(x) = (2^2-x^2)/( 2-x)}}}……….now we have the difference of square numbers and we can write it as {{{(2-x)(2+x)}}}

{{{f/g(x) = (2-x)(2+x)/( 2-x)}}}……….cancel {{{2-x}}}

{{{f/g(x) = 2 + x}}}……….solution


But, if you have {{{x}}}, then we will have:

{{{f(x) = 4-x}}}
{{{g(x) = 2-x}}}

Then: {{{f/g(x) = (4-x)/( 2-x)}}}………. this is a graphical form that simplifies to

{{{f/g(x) = (-x + 4)/( -x + 2)}}}……….