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

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)

Accepted Answer

Welcome back. I am so glad you're here. We are given two functions the F. Of X. Which is equal to the log of X And G of X. Which is equal to the negative log of X -1. Alright, we are to do several things with these two functions. We need to graph them on the same Cartesian plane. We need to identify their assam totes and we need to specify their domains and ranges. All right. We've got a lot to do. Let's start off sketching a rough Cartesian plane. We've got in a horizontal X axis of vertical Y axis. They come together at the origin in the middle. Now to graph our F of X. If you've got the log of X memorized, you know exactly what that looks like. That's great. Go ahead and graph that if you don't we can always plot a few points to see what's going on with log of X. And typically I would choose a few X values and figure out their corresponding Y values. But with a log I'm going to actually choose some Y values. Going to choose 10 and negative one and figure out their corresponding X values. Because we recall from previous lessons that when we have the log with the base B of X that equals Y. While at the same time we know that in exponential form be raised to the Y is equal to X. And in our given F of X. The B there isn't written, which means that B equals 10. It's a base 10. So for all of these we've got 10 raised to the Y. Which will equal R. X. And it's a little bit easier to plug in a value for Y. There and solve for X. So for the first one here we've got 10 raised to the instead of Y. It'll be 10 raised to the 1st 10 of the first. That equals 10. So our X. Here is 10. We look at our Cartesian planes starting in the origin 00. We're going to go up one for our Y. Value and then we're going to go to the right 10 for our X. Value. So that's approximately there. Alright. Our next point that we want to plot where we have Y equals zero. That will be 10 raised to the zeroth power. Anything raised to the zero is equal to one. So this is actually an X intercept will be zero for the Y. Value. We don't go up or down and we move one to the right for our X. Value. There's our X intercept. Now the last point we're going to plot for F of X. We've got 10 raise to the instead of why we have the negative first. Well that negative up in the exponent tells us that that 10 of the first is going to come down to the denominator. So this is 1/10 to the first which is equal to 1/10. So where Y equals negative one X equals 1/10. We look at our Cartesian plane. So we're going to go down one for our Y value and then go to the right to 1/10 for our X value. Now what's happening before and after? We have these three points, what's happening before or after? Well continuing to larger Y values going up with our Y values. X is going to get exponentially larger very fast. So that's just going to continue going in this direction. And for as we get smaller for our Y values as we get more and more negative, We're going to continue to have fractions but the power that the 10 is being raised, two in the denominator is going to get bigger and bigger. Those numbers are going to get closer and closer to zero but because they're fractions they will never actually reach zero. So we have what's called an assam tote as we recall from previous lessons right on the Y axis. So we have an assam tote where X equals zero. Okay, we got the grafton and the ASM totes done for F of X. Now let's work on G of X. Ok, we see two differences between F of X and G of X. One. Is this negative being multiplied in front and one is this minus one here in with the X in the parentheses. So what if each of those mean while this first one here, the negative out in front because that's not being applied directly to the exits, not inside of the parentheses, it's going to be a vertical change. And specifically because we're multiplying by a negative one, it is going to be a reflection over the X. Axis And the other change here. This -1 that is happening inside the parentheses. It's happening directly to the X. So that means it's going to be a horizontal change. And because we're subtracting one the change is that it's going to shift one to the right. It'll shift 1 to the right. Okay now to be very specific so that we choose the right answer choice. We can plot a few points here again so that we know exactly where things are going. So we have to do a little bit of work here to get this to change, we've got the negative log. Um the base is still 10 Of X -1 equal to Y. In order to get that into exponential form, we need to divide both sides by -1. So that the log isn't negative on the left. So now it's a log base 10. Um Excuse me not equals log base 10 of x minus one equals a negative y. And now we will take the 10 raised to the negative Y. And that equals x minus one. Switching to exponential form and we need to add one to both sides. So we have 10 raised to the negative y plus one equals X. We use those same Y values. Again we'll use +10 negative one. And now solving for X. So for Y equals one will put 10. Race to the negative instead of why? It's one plus one. Alright 10 raised to the negative one. Again that is 1/10. So we have 1/10 +11 is the same as 10/10. So that's 1/10 plus 20th. Which is 11 10th. So where Y equals one, X equals 11 10th. Which is very close to one. So let's go up one from the origin on our Cartesian plane. We're going up one for our Y. Value of one. And then for X. Value we're going to the right 11 10th which is just past one. Now let's sulfur Y. Equals 0 10 raised to the negative not why? But now zero plus one. Well 10 to the negative zero. That's 10 to the zero which is one. So we have one plus 11 plus one equals two. So our X intercept this time is at 20 on our Cartesian plane. We don't move up or down for Y. And we go to the right to for our X. Intercept now for negative one that will be 10 raised to the negative. Not why but negative one plus one. Well 10 raised to the negative negative one is 10 raised to the positive one which is 10 to the first which is 10. So we have 10 plus 1. 10 plus one equals 11. So where Y equals negative one. We go down one on our Cartesian plane from the origin for our Y value of negative one. We go to the right 11 for our X value. So now we have these three dots three points that we can connect and what's happening with them. Well for our um as why increases as the Y values get larger and larger again we're dealing with that um fraction. And as the Y values get larger and larger than that fraction is going to get smaller and smaller and closer this time because it has a plus one added to its going to get closer and closer to one but it will never actually reach one. So this time we've got an assam tote Where x equals one for our G. Of X. And as the Y values get smaller and smaller, those X values are quickly going to get bigger and bigger because we're dealing with exponents 10 raised to the bigger and bigger numbers. So that will quickly head to the right alright. We have got our graph sketched, we have identified our ASM. Totes. Now we just need to describe the domain and range. So let's separate our X. And our G. Of X a little bit. So for the domain and range for our F of X. Domain is all of the values of X. What's possible for our X values? Well, since X has an assam tote at zero it's going to have to be our X values are going to have to be greater than zero about our range. Those are the possible Y values. Why can be anything going from negative infinity to positive infinity. So that's all real numbers. Now let's take a look at our um domain and range for G of X. Ok for our domain, our X values, we've got an awesome toad at X equals one. So our X values are going to have to be greater than one. How about the range while the range the Y values, they could be anywhere from negative infinity to positive infinity. So again, this one is all real numbers. Okay, we've done it. We have solved all of the pieces of this and now we just need to figure out which one is the right answer choice. A really good place to look for the correct answer choice is keeping in mind our X intercepts for our F of X. The x intercept is going to be at 10 for our G of X. The X intercept is at 20. So those are two important things to keep in mind. Let's look at our answer choice. A. Well they've got the F of X intercept at but look at that G of X intercept that 0.20. So that one is not going to work. Alright, let's scroll down to answer choice A doesn't work. Now let's look at B. There affects well that they have that as the blue one here And that's an intercept at 0.20, nope B is not going to work either. Let's bring some of these this information down a little bit. Alright now we'll compare to C. And D. Alright so for C they've got intercepts at 10 for affects and 20 for G of X. That's great. And it looks kind of like our graph that we drew two. Alright let's look real quick at answer choice D. Well they have the G. Of X in red with the intercept at and that is not going to work. Our G intercept is at 20 for the X intercept. So answer choice C. Let's see if everything else works. Okay. The graph looks good as some totes for F of X at X equals zero. That's correct. For G of X it X equals one. That's correct. The domain of F of X. X is going to be greater than zero. That's correct. G of X is domain X is greater than one. That's correct. And for the range for F of X. All real numbers and the range for G of X. All real numbers. See works 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) = log x and g(x) = - log (x+3)

Key Concepts

Logarithmic Functions and Their Graphs

Transformations of Functions

Asymptotes and Domain/Range of Logarithmic Functions

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