Hey everyone. In this problem, we're asked to graph 2 different functions,
Now if I plug 1 into my function, that's
We know that it's going to increase really rapidly and get really steep there. Now for my negative values of x, we're going to see the same numbers but instead fractions as we had with positive 1, 2. So
I know that I'm going to get really steep on this side. And then on my left side here, I'm going to get really close to my x-axis but not quite cross it because we have an asymptote there. So I can go ahead and plot my asymptote as well. I know that it's at the x axis right at
Now remember that whenever we're dealing with inverse functions, we can simply swap our x and y values. So these values that I have for x over here are simply going to become my y values when working with my new function that is the inverse. And then my y values over here are going to become the x values of my inverse function. So let's go ahead and swap all of those points so that we can get them plotted on our graph. So starting with this point, I'm going to go ahead and switch that up.
So now it is 1,9 negative 2, just reversing my x and y. And then for negative one, one third, that becomes one third a negative one. 0,1 flip is 1,0. Then I have 3,1, which I flipped to get 1,3. And then my last point, flipping that 2 and that 9, I get 9,2.
Now let's go ahead and plot all of these points for our function
And then one ninth negative 2, we're going to be getting really close to our y-axis this time. So we can go ahead and connect all of these points to form our graph. And then we're getting really close to that y-axis, which tells us that we're dealing with a vertical asymptote up on this side right on our y-axis at
It can be all real numbers. Now, we know that our domain of our
So my range is simply going to go from my asymptote at that 0 point all the way to infinity. So my range here is simply from 0 to infinity. Now we know that our range is going to flip and become the domain of our other function of our inverse function, so that tells me that the domain of
Let me know if you have any questions.