How do you find the inverse of #f(x)=In(3-2x)+3#?

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Answer 1 · 2016-03-11T04:47:05

x = #( 3-e^(y-3))/2#, where y = f(x).

Using y = f (x ),

#e^(y-3)=e^(ln(3-2x))#

= #(3-2x)#
Rearrange for the inverse relation

#x = ( 3-e^(y-3))/2#

The graphs for both are one and the same.

Observe that x < 1.5.
graph{y - ln ( 3 - 2x )-3=0}

graph{x-1.5+0.5(2.718)^(y-3)=0}

Important for inverters:

For inverse of #y == abs f(x)#, use my explanation for the inverse

operator #(abs)^(-1)#.

#x = (abs)^(-1) (y)#, where

#(abs)^(-1) (y) = y, f(x) >= 0# and

#(abs)^(-1) (y) = -y, f(x) <= 0#.

For example, if #y = abs(x-1)#,

#x - 1 = (abs)^(-1)(y)#

# = y, x - 1 >= 0# and

#= -y, x - 1 <= 0#.