Question

Find the inverse function of f(x)= 1+squareroot of 1+2x

Answer 1

Answer:

Therefore the inverse function of  $f(x)=1+\sqrt{1+2x}$ is $\frac{x^2-2x}{2}$

Explanation:

We need to find the inverse of function $f(x)=1+\sqrt{1+2x}$

Function Inverse definition :

$\mathrm{If\:a\:function\:f\left(x\right)\:is\:mapping\:x\:to\:y,\:then\:the\:inverse\:functionof\:f\left(x\right)\:maps\:y\:back\:to\:x.}$

$y=1+\sqrt{1+2x}$$\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y$

$x=1+\sqrt{1+2y}$

$\mathrm{Solve}\:x=1+\sqrt{1+2y}\:\mathrm{for}\:y$

$\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}$

$1+\sqrt{1+2y}-1=x-1$

Simplify

$\sqrt{1+2y}=x-1$

$\mathrm{Square\:both\:sides}$

$\left(\sqrt{1+2y}\right)^2=\left(x-1\right)^2$

$\mathrm{Expand\:}\left(\sqrt{1+2y}\right)^2:\quad 1+2y$

$\mathrm{Expand\:}\left(x-1\right)^2:\quad x^2-2x+1$

$1+2y=x^2-2x+1$

$\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}$

$1+2y-1=x^2-2x+1-1$

$\mathrm{Simplify}$

$2y=x^2-2x$

$\mathrm{Divide\:both\:sides\:by\:}2$

$\frac{2y}{2}=\frac{x^2}{2}-\frac{2x}{2}$

$\mathrm{Simplify}$

$y=\frac{x^2-2x}{2}$

Therefore the inverse function of $f(x)=1+\sqrt{1+2x}$ is $\frac{x^2-2x}{2}$