1. ## Exponentials and Logs

Hi guys, I have come across this question that I'm not sure how to answer. It goes as follows

The function f is given by

f(x) = ln(3x-6) x>2

a)write down f^(-1)(x) - ie the reverse function

To do this, our teacher says to make f(x)=y, swap the x and ys, then solve for y

This gives me
y = ln(3x-6)
x = ln(3y-6)
e^x = 3y-6

isnt it then simply f^(-1)(x) = (e^x+6)/3

What do I do now??

The next question is b) write down the domain and range of f^(-1)(x), which I know seems really simple but im not too sure how to do it

Isn't the range for a function the domain for the inverse and vice versa?

Cheers

2. ## Re: Exponentials and Logs

You have done exactly right- that is the formula for $\displaystyle f^{-1}(x)$ and you don't need to do anything else. If you really want to you might 'simplify' that a little as $\displaystyle f^{-1}(x)= \frac{1}{3}e^x+ 2$.

Yes, the domain and range swap for the inverse function. Here, the domain of f is x> 2 (actually, that's the "natural domain". If [tex]x\le 2[tex] then $\displaystyle 3x- 6\le 0$ and ln(x) is only defined for x> 0) and the range is all real number (x very close to 2 will give 3x- 6 very close to 0 and so ln(3x- 6) "very close to $\displaystyle -\inty$" while if x is arbitrarily large, ln(3x- 6) is "very close to $\displaystyle \infty$". So the domain of $\displaystyle f^{-1}(x)$ is all real numbers and its range is y> 2.

3. ## Re: Exponentials and Logs

Good work thus far: $\displaystyle f^{-1}(x)=\frac{e^x+6}{3}$.
Clearly the domain is all real numbers and the range must be $\displaystyle (2,\infty)$. Think about it: $\displaystyle (\forall x)[e^x>0$.