Answer:
Therefore the inverse function of [tex]f(x)=1+\sqrt{1+2x}[/tex] is [tex]\frac{x^2-2x}{2}[/tex]
Explanation:
We need to find the inverse of function [tex]f(x)=1+\sqrt{1+2x}[/tex]
Function Inverse definition :
[tex]\mathrm{If\:a\:function\:f\left(x\right)\:is\:mapping\:x\:to\:y,\:then\:the\:inverse\:functionof\:f\left(x\right)\:maps\:y\:back\:to\:x.}[/tex]
[tex]y=1+\sqrt{1+2x}[/tex][tex]\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y[/tex]
[tex]x=1+\sqrt{1+2y}[/tex]
[tex]\mathrm{Solve}\:x=1+\sqrt{1+2y}\:\mathrm{for}\:y[/tex]
[tex]\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}[/tex]
[tex]1+\sqrt{1+2y}-1=x-1[/tex]
Simplify
[tex]\sqrt{1+2y}=x-1[/tex]
[tex]\mathrm{Square\:both\:sides}[/tex]
[tex]\left(\sqrt{1+2y}\right)^2=\left(x-1\right)^2[/tex]
[tex]\mathrm{Expand\:}\left(\sqrt{1+2y}\right)^2:\quad 1+2y[/tex]
[tex]\mathrm{Expand\:}\left(x-1\right)^2:\quad x^2-2x+1[/tex]
[tex]1+2y=x^2-2x+1[/tex]
[tex]\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}[/tex]
[tex]1+2y-1=x^2-2x+1-1[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]2y=x^2-2x[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}2[/tex]
[tex]\frac{2y}{2}=\frac{x^2}{2}-\frac{2x}{2}[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]y=\frac{x^2-2x}{2}[/tex]
Therefore the inverse function of [tex]f(x)=1+\sqrt{1+2x}[/tex] is [tex]\frac{x^2-2x}{2}[/tex]
