Question

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`?

Answer 1

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`