Question 1009988

Find the inverse of {{{f(x)}}}, where {{{f(x)=x^2+4x+3}}}

recall {{{f(x)=y}}}, so you have

{{{y=x^2+4x+3}}}........swap {{{x}}} and {{{y}}}

{{{x=y^2+4y+3}}}.............solve for {{{y}}}

{{{x=(y^2+4y+b^2)-b^2+3}}}..............complete square

{{{x=(y^2+4y+2^2)-2^2+3}}}

{{{x=(y+2)^2-4+3}}}

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

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

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

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

so, your inverse is

{{{f^-1(x)=(-2+- sqrt(x+1))}}}


{{{ graph( 600, 600, -10, 10, -10, 10, x^2+4x+3,sqrt(x+1)-2,-sqrt(x+1)-2) }}}