(a) What are the domain and range of $ f $?
(b) What is the x-intercept of the graph of $ f $?
(c) Sketch the graph of $ f $.
$ f(x) = \ln (x - 1) - 1 $
a) Domain: $x > 1$
Range: $(-\infty, \infty)$
b) $e+1 \approx 3.71$
you were given f of X, and we want to find its domain and range and X intercept and sketch the graph. So let's compare it to the basic function y equals the natural log of X. Hopefully, we know the shape of that curve. We know it has a vertical Lassen towed at X equals zero. We know it has an X intercept at one. We know it's domain is X is greater than zero. We can write that a zero to infinity if we want to. And we know it's ranges all real numbers. We can write that as negative infinity to infinity if we want to. Okay. And so how does that relate back to the graph we have? Okay, so the transformations that take place here we see we have a shift to the right one and we have a shift down one. Okay, now let's think about the domain so you can't have a negative or zero inside your locker them. And because of that, we need X minus one to be greater than zero. If X minus one is greater than zero, then X is greater than one. So we can write the domain as one to infinity when we shifted down one. It's still going to be going down to negative infinity and up to positive infinity. The range of all real numbers is not going to change. So the range is still all real numbers. To find the X intercept, we want to find the point with a Y coordinate of zero. So we substitute zero and for why we get zero equals the natural log of X minus one minus one. Let's add one to both sides and then remember, natural log means log Basie. So we have one equals log based E of X minus one. So if we rearrange this equation into its exponential form, we have e to the first power equals X minus one, and then we add one to both sides and we have e to the first, which is just e plus one equals X. So our X intercept is the point. He plus one comma zero. All right, now let's draw the graph and we have the benefit of having already looked at the graph of the natural log function. Remember, it had its vertical Assen toted X equals zero. We're gonna shift that to the right one. Okay. And now we also have the benefit of knowing one of the points on the graph e plus one comma zero Andy plus one is approximately 3.7. So we can plot that point and we can draw a graph.