Question 275302
let f(x)=3x-2x and g(f(x))=2x^2-3x+5. what is g(x)?
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I think you botched what {{{"f(x)"}}} equals because {{{3x-2x}}} is just {{{x}}},
and if {{{"f(x)"}}} were merely {{{x}}}, then {{{"g(x)"}}} would just be the same as
{{{"g(f(x))"}}} which would be {{{2x^2-3x+5}}}, and that would be a silly
problem.


So I will make up something else for {{{"f(x)"}}}, since I don't believe
you really meant {{{"f(x)"=3x-2x}}}. Suppose the problem were

let {{{"f(x)"=x-2}}} and {{{"g(f(x))"=2x^2-3x+5}}}. what is {{{"g(x)"}}}?

Then {{{"g(x)"}}} would have to be a quadratic function, say 

{{{"g(x)"=Ax^2+Bx+C}}}

Then 

{{{"g(f(x))"=A(x-2)^2+B(x-2)+C}}}

{{{"g(f(x))"=A(x^2-4x+4)+Bx-2B+C}}}

{{{"g(f(x))"=Ax^2-4Ax+4A+Bx-2B+C}}}

Rearrange the terms:

{{{"g(f(x))"=Ax^2+Bx-4Ax+4A-2B+C}}}

Factor x out of the 2nd and 3rd terms on the right

{{{"g(f(x))"=Ax^2+(B-4A)x+4A-2B+C}}}

Since {{{"g(f(x))"=2x^2-3x+5}}}, we equate coefficients:

{{{system(A=2,B-4A=-3,4A-2B+C=5)}}}

Substitute {{{A=2}}} into {{{B-4A=-3}}}

{{{B-4(2)=-3}}}
{{{B-8=-3}}}
{{{B=5}}}

Substitute {{{A=2}}} and {{{B=5}}} into

{{{4A-2B+C=5)}}}

{{{4(2)-2(5)+C=5)}}}

{{{8-10+C=5}}}

{{{-2+C=5}}}

{{{C=7}}}

So {{{"g(x)"=Ax^2+Bx+C}}}

becomes

 {{{g(x)=2x^2+5x+7}}}

Edwin</pre>