Question 1187397


If {{{f(x)=x+1}}} and {{{g(x)=x-1}}}

(a) 

{{{f(g(x))=f(x-1)}}}
{{{f(g(x))=(x-1)+1}}}
{{{f(g(x))=x-1+1}}}
{{{f(g(x))=x}}}


(b) 

{{{g(f(x))=g(x+1)}}}
{{{g(f(x))=(x+1)-1}}}
{{{g(f(x))=x+1-1}}}
{{{g(f(x))=x}}}

=> {{{f(g(x))=g(f(x))}}}


(c) 

Thus {{{g(x)}}}is called an  {{{inverse }}}function of{{{ f(x)}}}