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)
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