Question 919810

{{{f(x)=3x^3+2}}}, 

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

{{{f(x)=3x^3+2}}} .....since {{{f(x)=y}}} we have

{{{y=3x^3+2}}}...swap {{{x}}} and {{{y}}}

{{{x=3y^3+2}}} .........solve for {{{y}}}

{{{x-2=3y^3}}}

{{{(1/3)x-2/3=y^3}}}

{{{y=root(3,(1/3)x-2/3)}}} => {{{f^-1(x)}}}

{{{f^-1(x)=root(3,(1/3)x-2/3)}}}


now find the value of {{{x}}} so that {{{f^-1(x)=4}}}:


{{{4=root(3,(1/3)x-2/3)}}} ...both sides raise to power of {{{3}}}


{{{4^3=(root(3,(1/3)x-2/3))^3}}} 


{{{64=(1/3)x-2/3}}} ...solve for {{{x}}}

{{{64+2/3=(1/3)x}}}

{{{192/3+2/3=(1/3)x}}}

{{{194/3=x/3}}} ...if denominators same, then nominators are same too

{{{highlight(x=194)}}} 


check if {{{f^-1(x)=4}}}


{{{f^-1(194)=root(3,(1/3)194-2/3)}}}


{{{f^-1(194)=root(3,194/3-2/3)}}}

{{{f^-1(194)=root(3,192/3)}}}

{{{f^-1(194)=root(3,64)}}}

{{{f^-1(194)=root(3,4^3)}}}

{{{f^-1(194)=4}}}