by Integr88 » Thu 31 March, 2011 22:10
to inf the inverse of a function such as `y=8/3x – 2/3` we first solve for x
`y=(8/3)x – (2/3)` as we see `1/3` is common in both terms so factor it out
`y=1/3(8x-2)`
Multiply both side by 3 and we have:
`y*3=3*1/3(8x-2)` 3 on the right side cancels out and we have:
`3y=8x-2` now add 2 to both side of equation and we have:
`3y+2=8x-2+2`
`3y+2=8x`
We now get rid of x by multiplying both side of equation by `1/8` so the equation will look `(3y+2)*1/8=8x*1/8` and we have:
`(3y+2)/8=x`
now we replace x with y and we have: `(3x+2)/8=y`
we further simplify as `y=3/8x+2/8`
`:.` final inverse function is `y=3/8x+2/8`