Question 1206605


{{{f(x) = 4x − 1}}}

Verify that
{{{f}}} ∘ {{{f ^-1}}}
and

{{{f^-1}}} ∘ {{{f}}}

are both the identity function.


({{{f }}}∘ {{{f^-1}}}){{{(x) = f(f^-1(x))}}}
({{{f^-1}}} ∘ {{{f}}}){{{(x) = f^-1(f(x))}}}


find {{{f^-1}}}

{{{f(x) = 4x - 1}}}...{{{f(x) = y}}}

{{{y= 4x -1}}}...swap variables

{{{x= 4y -1}}}...solve for {{{y}}}

{{{x+1= 4y}}}

{{{y=x/4+1/4}}}

=>

{{{f^-1(x)=x/4+1/4}}}


({{{f }}}∘ {{{f^-1}}}){{{(x) =f(f^-1)=f(x/4+1/4)=4(x/4+1/4) - 1=x+1 - 1=x}}}

({{{f^-1}}} ∘ {{{f}}}){{{(x) =f^-1(4x-1)=(4x-1)/4+1/4=4x/4-1/4+1/4=x}}}


proven that
({{{f }}}∘ {{{f^-1}}}){{{(x) = f(f^-1(x))=x}}}
({{{f^-1}}} ∘ {{{f}}}){{{(x) = f^-1(f(x))=x}}}