How do you find f^-1x and how must x be restricted in f^-1x given f(x)=4+2cos(x-3) and 3<=x<=3+pi?

1 Answer
Jan 31, 2018

As with all inverse functions, start by taking equation y=f(x) and substituting x=f(y)

We have y = f(x) = 4 +2cos(x-3), hence we let: x = 4 + 2cos(y-3)

Doing some algebraic manipulation we get:

1/2(x-4) = cos(y-3)
:. y -3 = cos^-1(1/2(x-4))
:. y = cos^-1(1/2(x-4))+3

This is our f^-1, but first we must find the new domain.

Since dom f = [3, 3+pi] and the cosine function is one-to-one within any interval of pi, there is no need to restrict further. Now domf becomes ranf^-1.

The implied domain of f^-1 gives 1/2(x-4) in [0,pi], which implies that x in [4,2pi +4]

Hence:

f^-1: [4,2pi +4] rarr RR, f^-1(x) = cos^-1(1/2(x-4))+3