Question

(a) Find the domain off (x) = ln(ex − 3).
(b) Find F −1 and its domain.

Answer 1

Answer:

a) The domain of f(x) is $x > 1.1$.

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$

So we need

$e^{x} - 3 > 0$

$e^{x} > 3$

Applying ln to both sides

$\ln{e^{x}} > \ln{3}$

$x > 1.1$

So the domain of f(x) is $x > 1.1$.

(b) Find F −1 and its domain.

$F^{-1}$ 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

$x = \ln{e^{y} - 3}$

Here, we apply the exponential to both sides:

$e^{x} = e^{\ln{e^{y} - 3}}$

$e^{y} - 3 = e^{x}$

$e^{y} = e^{x} + 3$

Applying ln to both sides

$\ln{e^{y}} = \ln{e^{x} + 3}$

The inverse function is:

$y = \ln{e^{x} + 3}$

The domain is

$e^{x} + 3 > 0$

$e^{x} > -3$

$e^{x}$ 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.