Question 1021484


{{{f(x)=x^2+4}}}
{{{g(x)=x}}}
	
{{{f(g(x))=f(x)=x^2+4}}}

{{{f(g(1))=f(1)=1^2+4=5}}}

{{{f(2g(x))=f(2x)=(2x)^2+4=4x^2+4}}}


the inverse of {{{f(x)}}}:If f had an inverse, then its graph would be the reflection of the graph of f about the line {{{y = x}}}. 

since {{{f(x)=y}}} we have

{{{y=x^2+4}}}............swap {{{x}}} and {{{y}}}

{{{x=y^2+4}}}.........solve for {{{y}}}

{{{x-4=y^2}}}

{{{y=sqrt(x-4)}}}

so, the inverse is {{{f^-1(x)=sqrt(x-4)}}}

{{{ graph( 600, 600, -10, 10, -10, 10, x^2+4,-sqrt(x-4), sqrt(x-4)) }}}

 the inverse of {{{g(x)=x}}} is

{{{g^-1(x)=x}}}