f(x)= e^(3x)-k, where k is a constant that is greater than 0. what is f^-1(x)? what is the domain of f^-1(x) ?

Given that f(x) = e^(3x) -
k

We need to find the inverse function f^-1
(x)

Let f(x) = y

==> y=
e^(3x) - k

Now we will add k to both
sides.

==? y+k = e^(3x)

Now we
will apply the natural logarithm for both sides.

==>
ln (y+k) = ln e^3x

==> ln (y+k) = 3x ln
e

But ln e = 1

==> ln
(y+k) = 3x

Now we will divide by
3.

==> x= ln(y+k) /
3

Now we will rewrite x as
y.

==> y= ln (x+k) /
3

Then the inverse function is
:

f^-1 (x) = (1/3) * ln (x+k)

Now we will find the
domain.

We know that the domain is when (x+k) >
0

But we know that k>
0

Then the domain are x values such that x >
-k

Then the domain is x E ( -k,
inf)