# Find the inverse of the function below. Graph the function below and the inverse function.`f(x)= 2ln(3 -x) +6`

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To solve this problem, and others like it, we need to switch `x` and `f(x)`. What this will do for us is **switch the inputs and outputs**, which is the essence of finding the inverse function! We are looking at which input produces which output by now starting with the output!

So, let's switch up `x` and `f(x)`. However, instead of `f(x)`, let's use `f^-1(x)`:

`f(x) = 2ln(3-x) + 6`

`x = 2ln(3-f^-1(x)) + 6`

Now, we solve this equation like any other! Let's start by subtracting 6 from both sides:

`x-6 = 2ln(3-f^-1(x))`

Now, divide both sides by 2:

`(x-6)/2 = ln(3-f^-1(x))`

To get rid of the natural log, we make both sides exponents of `e`. This action cancels out the natural log and allows us to get inside that operation:

`e^((x-6)/2) = 3-f^-1(x)`

Now, we subtract 3:

`e^((x-6)/2) - 3 = -f^-1(x)`

Finally, we multiply both sides by -1 to get our final inverse function:

`f^-1(x) = 3-e^((x-6)/2)`

**There you have it! Your inverse function is an exponential function,** which makes sense because our starting function was a logarithmic function. If you want to think about domains and ranges, the domain and range of the inverse function are the range and domain of the starting function, respectively. In other words, domain and range switch roles, just like how `x` and `f(x)` switched roles in our inverse function.

Something else interesting about this is how these functions look when graphed. All inverse functions have an interesting line of symmetry at the line `y=x`. This fact is easily seen when you consider how we found the inverse by effectively switching the x and y values for all coordinates on the original graph. Let's see how it worked for our function:

The blue graph is our original logarithmic function, the red graph is our inverse function, and the black line is the line `y=x`. It isn't exactly clear on this graph, but the lines are reflections of each other across that line. The graph doesn't capture this well, but the first function actually has an **asymptote **at `x=3`, and it will proceed down near-vertically when it gets near `x=3`. The range of `f(x)` is, in fact, all of the real numbers. This fact is reflected in the domain of the inverse, which is clearly all real numbers because it is an exponential function. Notice that the graph of `f^-1(x)` has a similar asymptote at `y=3`, reflecting the original function's asymptote at `x=3`.

Keep in mind, this method works well **as long as the function is 1 to 1**, meaning that every `x` in the domain is mapped to one and only one `y`. This condition is important when you consider that the inverses of functions that do not meet this test, such as parabolas, are not functions because they do not fulfill the vertical line test.

I hope this all helps!