Question 1044096
If {{{f(x) = 5x}}} and {{{g(x)=2x^2-1}}}

How would you set up:
{{{(fog)(x)}}}=>
{{{(f)(x)*(g)(x) = (5x)(2x^2-1)=10x^3-5x}}}

{{{(gof)(x)}}} 

 {{{(gof)(x)=g(f(x))= 2(5x)^2-1=2*25x^2-1=50x^2-1}}}

{{{g^-1(x)}}} 
recall that
{{{g(x)=y}}}
{{{y=2x^2-1}}}...to find inverse, switch {{{x}}} and {{{y}}} first

{{{x=2y^2-1}}}..........solve for {{{y}}}

{{{x+1=2y^2}}}

{{{(x+1)/2=y^2}}}


{{{y=sqrt((x+1)/2)}}}

{{{y=sqrt((x+1))/sqrt(2)}}}

so, {{{g^-1(x)=sqrt((x+1))/sqrt(2)}}} or {{{g^-1(x)=-sqrt((x+1))/sqrt(2)}}}


{{{f^-1(x)}}} ?

{{{f(x) = 5x}}}
{{{y=5x}}}
{{{x=5y}}}

{{{y=x/5}}}=>{{{f^-1(x)=x/5}}}