Question 88250
{{{fog(x)=f(g(x))}}}


{{{f(g(x))=f(x^2-3x)}}} F of G of x is equivalent to F of {{{x^2-3x}}}


{{{f(x^2-3x)=3(x^2-3x)-2}}} Now plug in {{{x=x^2-3x}}} into {{{f(x)}}}


{{{f(x^2-3x)=3x^2-3(3x)-2}}} Distribute


{{{f(x^2-3x)=3x^2-9x-2}}} Multiply


So {{{f(g(x))}}} is a function