Answer:
a) The domain of f(x) is
.
b)
The inverse function is:
![y = \ln{(e^{x} + 3)}The domain is all the real values of x.Step-by-step explanation:(a) Find the domain off f(x) = ln(e^x − 3)The domain of f(x) = ln(g(x)) is g(x) > 0. That means that the ln function only exists for positive values.So, here we have[tex]g(x) = e^{x} - 3](https://tex.z-dn.net/?f=y%20%3D%20%5Cln%7B%28e%5E%7Bx%7D%20%2B%203%29%7D%3C%2Fp%3E%3Cp%3EThe%20domain%20is%20all%20the%20real%20values%20of%20x.%3C%2Fp%3E%3Cp%3E%3Cstrong%3EStep-by-step%20explanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%28a%29%20Find%20the%20domain%20off%20f%28x%29%20%3D%20ln%28e%5Ex%20%E2%88%92%203%29%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EThe%20domain%20of%20f%28x%29%20%3D%20ln%28g%28x%29%29%20is%20g%28x%29%20%3E%200.%20That%20means%20that%20the%20ln%20function%20only%20exists%20for%20positive%20values.%3C%2Fp%3E%3Cp%3ESo%2C%20here%20we%20have%3C%2Fp%3E%3Cp%3E%5Btex%5Dg%28x%29%20%3D%20e%5E%7Bx%7D%20-%203)
So we need


Applying ln to both sides


So the domain of f(x) is
.
(b) Find F −1 and its domain.
is the inverse function of f.
How do we find the inverse function?
To find the inverse equation, we change y with x to form the new equation, and then we isolate y in the new equation. So:
Original equation:
f(x) = y = \ln{e^{x} - 3}
New equation

Here, we apply the exponential to both sides:



Applying ln to both sides

The inverse function is:

The domain is


is always a positive number, so it is always going to be larger than -3 no matter the value of x. So the domain are all the real values.