Stuck on the inverse of a function?

Question

What is the inverse of f(x)=4cos(180(x-2))+1 with the restriction on the domain 2<x<3

And what is the inverse of f(x)=1+tan(90x) with the restriction on the domain -1<x<1

Answer 1

(1).

The function is f(x) = 4cos[180(x-2)]+1 and the domain is 2<x<3.

Replace y by f(x) and solve for x.

y = 4cos[180(x-2)]+1

y - 1 = 4cos[180(x-2)]

(y - 1)/4 = cos[180(x-2)]

cos^-1[(y - 1)/4] = 180(x-2)

{cos^-1[(y - 1)/4]}/180 = x-2

{cos^-1[(y - 1)/4]}/180 + 2 = x

x = {cos^-1[(y - 1)/4]}/180 + 2

Interchange x and y.

y = {cos^-1[(x - 1)/4]}/180 + 2

Replace y by f ^-1(x).

f ^-1(x) = {cos^-1[(x - 1)/4]}/180 + 2.

Answer 2

(2).

The function is f(x) = 1+tan(90x) and the domain is -1<x<1.

Replace f(x) by y and solve for x.

y = 1+tan(90x)

y - 1 = tan(90x)

tan^-1(y - 1) = 90x

(1/90) * tan^-1(y - 1) = x

x = (1/90) * tan^-1(y - 1)

Interchange x and y.

y = (1/90) * tan^-1(x - 1)

Replace y by f ^-1(x).

f ^-1(x) = [tan^-1(x - 1)]/90.