Find the inverse

Question

y=1+In(-x) find the domain range and asymptote

Answer 1

If you are interested in finding the domain, range and asymptote for the graph of

y = 1 + ln(-x)

You should learn how to sketch the graph of these functions. If you don't know what the graph of y = ln(x) looks like, grab a graphing calculator or google a graphing calculator (Desmos is a good one) and learn this graph.

The reason why is because, if you have

y = ln(-x) the graph will flip over the y - axis. This happens for any function of the form y = f(-x)

Try experimenting on the graphing calculator, try testing out

y = x³ -----vs----- y = (-x)³

y = e^x------vs-----y = e^(-x)

y = √x------vs-----y = √(-x)

In all of these examples, the original graph should be reflected over the y - axis.

For the graph of y = 1 + ln(-x)

the graph of y = ln(-x) gets shifted up by 1 unit.

So, anything of the form y = f(x) + b moves up/down by "b" units. Test out a couple functions, play around and experiment, so this makes sense to you.

Regardless, if you look at the sketch of y = 1 + ln(-x) you will notice that the graph is all on the left hand side, except it doesn't touch x = 0. Thus, your domain is

D: {x: x < 0}

As for the range, if you look carefully at the graph, you'll notice that the graph continues to grow as you move to the left, (I.e. the graph will hit all the y - values.) Thus,

R: all real numbers

As for the asymptote, there is only a vertical asymptote, as you can see from the graph. The graph is getting closer to x = 0, but will never touch it, thus, the graph has a

Vertical Asymptote at x = 0

There is no Horizontal asymptote, because the graph doesn't "level off" to a certain number. An example of a graph that has a horizontal asymptote is y = e^x (It has a horizontal asymptote at y = 0, because the graph starts to level off toward y = 0 as you move more toward the left).

Hope this helps

Mr. K