What is the inverse of #f(x) = 2ln(x-1)-ix-x#?

1 Answer
Jul 5, 2017

#f^(-1)(x)=e^(ix)-x+1##

Explanation:

If f"#f(x)=2"ln"(x+1)-ix-x#, then #f^(-1)(x)#is the inverse, or the reflection of #f(x)# in the line #y=x#.

So, let's make #f(x)=2"ln"(x+1)-ix-x=0#, then #2"ln"(x+1)=ix+x=i(2x)#, dividing both sides by 2 gives #f(x)="ln"(x+1)=(i(2x))/2=ix#.

As #e^("ln"(a))=a#, #e^(ln(x+1))=x+1=e^(ix)#.

Now take away #(x+1)# from both sides to make it equal 0, #0=e^(ix)-x+1#.