# Math Help - Help with inverse function

1. ## Help with inverse function

For my precalculus 12 assignment, I have to determine the inverse function of f(x)=3^x-1 -2. I'm not sure how. Could someone point me in the right direction?

2. ## Re: Help with inverse function

Originally Posted by Han
For my precalculus 12 assignment, I have to determine the inverse function of f(x)=3^x-1 -2. I'm not sure how. Could someone point me in the right direction?
Please use parenthesis: Is your function $f(x) = 3^{x - 1} - 2$

The main idea is to solve for x. So
$y = f(x) = 3^{x - 1} - 2$

$y + 2 = 3^{x - 1}$

How can you apply a logarithm here?

-Dan

3. ## Re: Help with inverse function

Originally Posted by Han
For my precalculus 12 assignment, I have to determine the inverse function of f(x)=3^x-1 -2. I'm not sure how. Could someone point me in the right direction?
you're going to have to write f(x) unambiguously using parentheses. From what you've written I can't tell what f(x) is.

is it $f(x) = 3^{x-1} - 2$ ?

if so let

$y= 3^{x-1} - 2$

solve for $x$. Then in your resulting formula for $x$ replace $x$ with $f^{-1}(x)$ and replace $y$ with $x$ and you'll be done.
This method works for whatever $f(x)$ is.

4. ## Re: Help with inverse function

Apologies for the lack of parenthesis, but yes that was the equation I was trying to present. Thanks.

5. ## Re: Help with inverse function

A point about notation: If the original function is given as a function of "x": f(x), then the inverse function will typically also be given as a function of x.

For example, if you are given function f(x)= 4x- 1 then to find the inverse function write y= 4x- 1 and solve for x: 4x= y+ 1 so x= (y+ 1)/4. So you would write the inverse function as $f^{-1}(x)= (x+ 1)/4$.

6. ## Re: Help with inverse function

Originally Posted by Han
For my precalculus 12 assignment, I have to determine the inverse function of f(x)=3^x-1 -2. I'm not sure how. Could someone point me in the right direction?
I would rewrite it as $x=3^{y-1}-2$.
Then $3^{y-1}=x+2\\y-1=\log_3(x+2)\\y=\log_3(x+2)+1$

-\$

7. ## Re: Help with inverse function

He then solved for y. That is equivalent to first solving for x and then writing this as a function of x rather than y as I noted in my previous post.