Graph the solution set of each system of inequalities.

Question

$y\le\log x$

$y\ge\left|x-2\right|$

Accepted Answer

Welcome back everyone in this problem for the system of inequalities where Y is less than or equal to the log of negative X and Y is greater than or equal to the absolute value of X plus three minus two. We want to graph its solution set here. We have our enter choices possible solutions for A B see and D says that there is no solution set. And now below that, I've put my own graph that will help us to plot our own system of inequalities. And that's important because if we can plot our inequalities, share the relevant regions, then wherever our regions intersect is going to give us our solution set. So let's go ahead and try to do that. Let's start with Y is less than or equal to the log of negative X. OK. Now, here we know that since we are dealing with a logarithm, it's going to be an asymptotic function. And in this case, our graph will be asymptotic to the Y axis. So let's try to get an idea of what the points will look like and that will help us to plot OK. Now we know that the term inside the logarithm has to be positive. So the only values of X that are possible are going to be negative. OK? Because once X is negative and negative negative value will make this term inside the log positive. So let's let Y OK be equal to the logarithm of negative X. And as we said, we know that the domain of this logarithmic function is where X is less than zero. Now we could do a quick table just to get an idea of what these values will look like. OK? So let's say here that we have a table of X and Y values. OK. And let's start from, let's start from X being equal to negative five all the way up to negative one. OK? Because remember as we said, X has to be a negative value. No, when X is negative five, OK, then we should get that Y is 0.699 when X is negative four, Y is 0.602. And again, we're getting these values by simply putting it into our calculator. So for example, when X equals negative three, then we're finding why as the logarithm of negativity and the log of negative three is 0.477 when X is negative two, Y is 0.301. And when X is negative one, Y equals zero, now if we plot our points, OK, if we try to plot our points on a graph, as we said, uh for negative 50.669 negative 40.602 negative 30.40 S 0.477 negative 20.301 negative one. We should get it to be zero. And then as we say, it's asymptotic. So if we were to go ahead and plot here, OK, then our graph would probably look something like this. OK. It's asymptotic to the Y axis. And let me just try to, this curve isn't going to be exact actually. So you know what, let me just make sure I draw as smooth a curve as I can. OK? So you know, it's just gonna be asymptotic. It's when you get close to the X axis but never becoming an X value. Now, which region are we going to share to satisfy the inequality? Let's test it. OK. Let's test, let's choose a random point to test. So let's test it with the point negative 11, given the condition that X must be less than zero. So let's test it here. If this point satisfies the inequality, we're going to share the region above the line. But if not, we're going to share the region below the line. OK. Now for negative 11, the question we're asking ourselves is if one is less than or equal to the log of negative one multiplied by negative one. No, is that the case in other words, the log of negative negative one, that's what I should have written here is that the case does it satisfy? Well, the log of negative negative one is equal to the log of one which is zero. OK? And is one less than or equal to zero. It is not. So, since the inequalities false, the test point is not in the shaded region which means that the shaded region must be below the line in the other direction. This must be our shaded region. OK? So let's just draw our shaded region here. OK. So this is the shaded region for Y is less than equal to the log of, of negative X. Now let's try our next uh inequality. OK. Let me put that in blue here. So the next inequality is Y is greater than or equal to the absolute value of X plus three minus two. Now how are we going to figure out what values these are? Well, we know that we could rewrite, OK. We are sorry, we could let our equation be Y equals the magnitude of X plus three OK minus two, which we can rewrite as the magnitude of X minus negative three, OK minus two. And we do that because no, when we write it in this format, this graph is the horizontal translation of three units left and the vertical translation of two units down of the parent function Y equals the absolute value of X OK. Now if we think about it here for our graph, if we could go ahead and plot a few points, then we should be able to solve, we can plot some points from Y equals the absolute value of X. Then we should be able to do our graph here. OK. Now, for Y being equal to the absolute value of X, OK, then at X equals zero, Y is going to be equal to the absolute value of zero, which is zero at X equals one, Y is the absolute value of one which equals one and at X equals negative one, Y is the absolute value of negative one which is positive one. So for Y equals the absolute value of X, we have the points 0011 and the negative 11 on that graph. So if we were to graph that here, that basic value here, OK? Let's just take this one out, then we're going to have our line going through here. OK? And we're gonna have our line going in the other direction. OK? And now we have our graph for Y equals uh the absolute value of X. Now, as we said, this graph is going to be a shift and this shift if we go ahead and do the shift, it's going to be three units to the left. So 123, OK? And two units down 12. OK. So now we can tell that this graph is going to cover this region. And again, let me just go ahead and extend it uh for the purpose of this problem. OK. So now that we know our graph is here, OK, then we need to figure out which region to shade is it going to be below our graph or above our graph. Let's use the test point of 00. OK. So now testing it with 00, the question we're asking ourselves is zero greater than or equal to the absolute value of zero plus three minus two. Well, zero plus three is three. The absolute value of three is three and three minus two equals one. So is zero greater than or equal to one. No, it's not. Therefore, since the inequality is false, that means that point does not reside in the region that satisfies the inequality. Therefore, the shaded region must be above the line we need to share above the line. So now if we go ahead and do that, OK, let's go ahead here. So we have this part is shaded, then this part is shaded and now we're gonna shade everything above the line. OK? Not this shaded region in blue is going to satisfy the inequality Y is greater than or equal to the absolute value of X plus three minus two. Know that we have both inequalities which region is the solution set. Well, this is one of the reasons why I like to keep the numbers because in this, keep the colors, sorry, because we can tell which regions intersect where our colors overlap. Now, if you take a closer look notice here that in this region in this region that I'm about to shade in black, this is the region where our colors overlap or that means that's where satisfies both inequalities. And thus this region represents the solution set. If we go back to our answer choices or graphs for our answers and compare, we can tell them that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Graph the solution set of each system of inequalities.﻿y≤log⁡xy\le\log xy≤logx﻿﻿y≥∣x−2∣y\ge\left|x-2\right|y≥∣x−2∣﻿

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