Question

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

Answer 1

Therefore the inverse function of
`f(x)=1+√(1+2x)` is
`(x^2-2x)/(2)`

Step-by-step explanation:

We need to find the inverse of function
`f(x)=1+√(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+√(1+2x)`
`\mathrm{Interchange\:the\:variables}\:x\:\mathrm{and}\:y`

`x=1+√(1+2y)`

`\mathrm{Solve}\:x=1+√(1+2y)\:\mathrm{for}\:y`

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

`1+√(1+2y)-1=x-1`

Simplify

`√(1+2y)=x-1`

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

`\left(√(1+2y)\right)^2=\left(x-1\right)^2`

`\mathrm{Expand\:}\left(√(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`

`(2y)/(2)=(x^2)/(2)-(2x)/(2)`

`\mathrm{Simplify}`

`y=(x^2-2x)/(2)`

Therefore the inverse function of
`f(x)=1+√(1+2x)` is
`(x^2-2x)/(2)`