# 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)...

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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)

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