In Exercises 39–40, graph f and g in the same rectangular coordin...

Question

In Exercises 39–40, graph f and g in the same rectangular coordinate system. Use transformations of the graph of f to obtain the graph of g. Graph and give equations of all asymptotes. Use the graphs to determine each function's domain and range. f(x) = ln x and g(x) = - ln (2x)

Graph of f(x) = ln x (red) and g(x) = -ln(2x) (blue) with asymptotes and key points.

Graph of g(x) = -ln(2x) (red) and f(x) = ln x (blue) with transformations and key points.

Graph of f(x) = ln x (red) and g(x) = -ln(2x) (blue) with asymptotes and key points.

Graph of g(x) = -ln(2x) (red) and f(x) = ln x (blue) with transformations and key points.

Accepted Answer

Welcome back. I am so glad you're here. We are given two functions F. Of X equals log base E. Of X and G of X. Which equals the negative natural log of five X. And we're to do several things. We need to graph both of these on the same Cartesian plane. We need to determine their assam totes and we need to identify their domains and ranges. So let's start off with the graph. We'll draw a rough sketch of a Cartesian plane. We've got a vertical Y axis and a horizontal X axis. They meet at the middle in the origin. And to graph our first R. F. Of X equals log base A. Of X. If you know that one if you've got it memorized, great if you don't you can always plot a few points to see what's going on. And normally I would plug in a few X values and solve for Y. But when we're dealing with a log that's a little bit easier to do. If we plug in some Y values, I'm going to plug in 10 and negative one for Y values. And solve for X. And here's why when we have log base E. Of X equal to Y. Which is what we have with our F. Of X. We have the same thing as X equals E. R. Base raised to the Y. Which we recall from previous lessons. So now if I plug in some Y values it'll just be E raised to that Y. And that will give us our X values. So one other thing to recall. E. Is a constant term. E. Is that natural number and it's equal to approximately not exactly 2.718. And then it keeps going And we're just going to use 2. here. So plugging in one for why? We've got E raised to the first power which is we know E. Is 2.718 raised to the first power. So our X value here is 2.718 raised to the first power which is 2.718. Great. We've got one point that we can plot. So looking at our Cartesian plane we go over approximately 2.718 to the right on our X axis and then go up one on our Y axis. So there's one point Alright now for our Y value of zero that will be E raised to the zero. Anything raised to the zero is equal to one. So there's our x intercept where X is equal to one, then y is equal to zero. So right here on our Cartesian plane we go one to the right for X and don't go up or down for our Y value. Now let's platform Y equals negative one. That will be E raised to the negative one. And since we have a negative one in our exponents we are going to put that E raised to the first in the denominator. So instead of E. That's 1/2 10.718 raised to the 1st and 2.718 raised to the first is 2.718. So this is 1/2 0.718 which is approximately equal to one half. So where we've got negative one on our Y axis going down one from the origin that will put us at approximately one half for our X value. Alright now we've got these three points that we can connect but what's going on before and after? Well as our Y values go up. So above one, that's going to be our E. Value. Race to higher and higher exponents or 2.718. Race to the second to the third to the fourth. That's going to get to be pretty big. X values pretty quickly. So this is going to go up to the X values pretty quickly as our Y values increased slowly. And how about as our Y values decrease while that negative one, negative two negative three etcetera. Those are going to be fractions and we're going to have bigger and bigger numbers in the denominator as our Y values get bigger and bigger. So these numbers are going to get closer and closer and closer to zero for our X values as our Y values increase. But because they are fractions, they will never quite reach zero. And this goes to our second thing that we're working on here. The ASM totes because our values are getting closer and closer to zero, but we'll never actually reach zero. We're going to have an ASM tote Where x equals zero for our F. Of X. Alright, so we've got that graft are fx's graft and we've identified it as um tote. Now let's look at our G. Of X. G of X is very similar. It's using the natural log and it's the negative natural log of five X. So there's two differences here between our log base E. Of X and our negative natural log of five X. This negative in front because that's being applied in front, not directly to our X. That implies that we're going to have a vertical change And the change that's occurring here because it's being multiplied by -1. Is that we are going to have reflection over the X axes. And what's our second change going to be here. Alright, this five is being multiplied by the X. Inside the parentheses there, that means we're dealing with a horizontal change because it's happening directly to that X. And because it's being multiplied by a number that is greater than one, it is going to indicate horizontal shrinking and now to get the most precise values so that we can determine which ones are are appropriate graphs as we're looking at our answer choices. I'd also like to graph a few points here for X and Y values and we can again, graph 10 and negative one for our Y values. We're still dealing with logs. So it's easier to graph are Y values first. And or to identify our Y values. And then from that figure out our X values. So in order to do this we'll need to do a little bit of rearranging. We've got that Y equals that's our G. Of X equals the negative natural log of five X. Another way to say this is that we've got Y equals the negative log base E. Of five X. And so it will translate again to E raised to the Y. But before we can start doing that we're going to need to divide both sides by negative one in order to get rid of that negative in front of the log based E. So we are dealing with a negative Y over on the left equals R log base E. Of five X. And so now translating that into exponential form, it will be the E raised to the negative Y. And that will equal not X. But now it will equal five X. So anything to get our X by itself, we're going to need to divide by five. So anything we would have will be divided by five. So to solve for X, it's going to be where X equals E raised to the negative Y divided by five. Okay, we can do this plugging in for Y equals one will have E raised to the negative one divided by five. E. Raised to the negative one. Well that means the E. Is going to go down into our denominator. So this is going to be equal to one over five. E to the first. Well five E to the first, that's just the 2.718 times five. So this will equal putting it into a calculator approximately 0.7. So graphing on our Cartesian plane where Y equals one. Where we go up one from the origin for our Y value X equals 10.7. A very small number. So that's approximately here on our Cartesian plane. Now let's solve for Y equals zero. So this time we're putting in E raised to the negative zero, divided by five. Okay, well erased to the negative zero. That's going to be E 20, which is just one. So this equals 1/5. Or putting that in a decimal? Like the other ones. That's 0.2. So graphing that on our Cartesian plane for y equals zero. We won't be moving up or down. We're on the X axis and the X value is at 0.2. So just a little bit higher than our 0.7. But really not much. Not even all the way up to that one. Alright, how about for y equals negative one. Okay, this is E raised to the negative negative one over five. Okay, well the negative negative one. That's a positive one. So this is E to the first over five. So that's that 2.718 divided by five. Putting that into a calculator equals approximately 0.54. So where Y equals negative one. We've got a 0.54 to the right on our X axis which is just slightly about here. So what's going on after we connect these three lines? Well are 0.7 value. Remember this other one was flipped over the X axis and then it shrank horizontally. So as Y gets larger this time our X values are going to get smaller and smaller closer to zero and this time as why becomes more and more negative. Our X values are going to increase. So this is what our graph should look like. Okay we got part one done now the ASM totes we already identified the ASM tote for F of X. The G of X is going to be the same thing. X is going to approach zero but it will never quite reach it. So we've got our ASM totes identified now for the domain and range. Okay, the domain are all of the possible X values that we're dealing with and X is going to have to be greater than zero because of that. ASM tote at X equals zero for both of these two functions. Now. How about the range? That's all the possible. Why values? Well the Y values could be anything from negative infinity. Up to positive infinity. So it could be all real numbers for both of our functions. Okay. We have done it now. We just have to figure out which of these answer choices. Works. Okay. Looking at a that graph that looks pretty good when we're looking at graphs. A good place to look and see if we've got the right answers is at our intercepts and this time our X intercepts because there are no Y intercepts because of our ASM tote. So our X intercept here for our F of X is at 10 and yeah, that's what we've got on this graph and our X intercept for G of X is at 0.20. And yeah, that's what we've got for our G of X on part A. So this graph looks pretty good. As some totes at X equals zero for both F of X and G of X. Great, that's correct. And for our domain, well, X is greater than zero for both of these. Yes, that is correct. Range all real numbers for both functions. Okay. A is looking pretty good. Let's do a really quick check to see what's going on with these other ones to see if maybe we've missed something and we're going to do a quick check to looking at our F of X is going to have that X intercept at 10 and r G of X is going to have that intercept at 0.20. That will be a good clue. Okay now scrolling down a little bit so that we can see B. Um X intercept for F. Of X at 10. Look they used different colors on this one even they put the F. Of X as the blue and no that one doesn't have 0. intercept. So no that one's not going to work. All right let's try C. Have they got 10 for their affects? Yes it does. That's good. But the G. Of X. Is at 20 not at 0.20. So C. Does not work now let's check out our D answer. All right. They've got that's that same thing as the last one. It has the 10 for affects but it has 20 for oh that's actually their affects. They switched to the F. Of X. And G. Of X. For those two. So that one also doesn't work. All right. So a does work for our answer for this one. Well done. We'll catch you on the next one

In Exercises 39–40, graph f and g in the same rectangular coordinate system. Use transformations of the graph of f to obtain the graph of g. Graph and give equations of all asymptotes. Use the graphs to determine each function's domain and range. f(x) = ln x and g(x) = - ln (2x)

Key Concepts

Logarithmic Functions

Transformations of Functions

Asymptotes

Watch next