F(x) = 4tan(2x-2) How do I find the inverse, image & domain

If f(x) = 4tan(2x-2) and if f:(-pi/2, pi/2)-->R

How do I find the inverse, image & domain?

I would appreciate any help and explanation - thank you.

The inverse function (f^-1(x)) of a given function (f(x)) must have the following properties:

f^-1(f(x)) = x

and

f(f^-1(x)) = x

First, let's find the domain by asserting the two following inequalities:

-π/2 < 2x - 2 < π/2

Add 2:

2 - π/2 < 2x < π/2 + 2

divide by two:

1 - π/4 < x < π/4 + 1

One way of discovering the inverse function is to let x = f(y) and then solve for y.

x = (4)tan(2y - 2)

x/4 = tan(2y - 2)

tan^-1(x/4) = 2y - 2

tan^-1(x/4) + 2 = 2y

y(x) = (1/2)tan^-1(x/4) + 1

We must now check that y(x) is f^-1(x) by doing the two tests:

y(f(x)) = (1/2)tan^-1({(4)tan(2x - 2)}/4) + 1

y(f(x)) = (1/2)tan^-1({tan(2x - 2)}) + 1

y(f(x)) = (1/2)(2x - 2)) + 1

y(f(x)) = x - 1 + 1

y(f(x)) = x

Now evaluate f(y(x))

f(y(x)) = 4tan(2{(1/2)tan^-1(x/4) + 1} - 2)

f(y(x)) = 4tan(tan^-1(x/4) + 2 - 2)

f(y(x)) = 4tan(tan^-1(x/4))

f(y(x)) = 4(x/4)

f(y(x)) = x

Now let's check the domain:

y(1 - π/4) = (1/2)tan^-1({1 - π/4}/4) + 1

y(1 - π/4) = (1/2)tan^-1({1/4 - π/16}) + 1

y(1 - π/4) ≈ 1.05

y(1 + π/4) = (1/2)tan^-1({1/4 + π/16}) + 1

y(1 + π/4) ≈ 1.21

y(x) is defined over the domain.

Therefore, f^-1(x) = (1/2)tan^-1(x/4) + 1

You are welcome.

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Domain for tan(x) ; -pi/2 < x < pi/2
So -pi/2 < 2x-2 < pi/2, => -pi/4 +1 < x < pi/4 +1 Thus domain is the interval (-pi/4 +1, pi/4+1 )
Image of f is the full real line R.

Let y = 4tan (2x-2)

tan^(-1) (y/4) = 2x -2

x= 1 +1/2 tan^-1 (y/4)

So f^-1 (x) = 1 + 1/2 tan^-1 (x/4) is the inverse function