Question 1180065


{{{g(x) = 3x+4}}}

 determine {{{n}}} such that {{{g(-3) = g^-1(n)}}}

first find {{{g^-1(x)}}}

{{{g(x) = 3x+4}}}......replace {{{g(x) }}} with {{{y}}}

{{{y= 3x+4}}}.....swap variables

{{{x= 3y+4}}}........solve for {{{y}}}

{{{x-4= 3y}}}

{{{y=x/3-4/3}}}

=> {{{g^-1(x)=x/3-4/3}}}=> it will be {{{g^-1(x)= g^-1(n)}}}


now find {{{g(-3) }}}

{{{g(-3) = 3(-3)+4}}}

{{{g(-3) = -9+4}}}

{{{g(-3) = -5}}}

then find
{{{g^-1(n)=-5}}}

{{{-5=n/3-4/3}}}

{{{-5+4/3=n/3}}}...both sides multiply by {{{3}}}

{{{-15+4=n}}}

{{{n=-11}}}