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At the point shown on the function above; which of the following Is true? 0f' < 0,f"' > 0 of' < 0,f'= <0 0f' > 0,f"...

Question

At the point shown on the function above; which of the following Is true? 0f' < 0,f"' > 0 of' < 0,f'= <0 0f' > 0,f" < 0 0f' >0,f" >0

At the point shown on the function above; which of the following Is true? 0f' < 0,f"' > 0 of' < 0,f'= <0 0f' > 0,f" < 0 0f' >0,f" >0



Answers

Multiple Choice If $a$ is a nonnegative real number, then which inequality statement best describes $a ?$ (a) $a<0$ (b) $a>0$ (c) $a \leq 0$ (d) $a \geq 0$

From able options. These the correct option that is a piece greater than are equal toe zero and given close in e ease, non negative real member. That means either it could be zero are more than that.

If two is less than X is less than six. Which of the statements about X are necessarily true, in which art. Okay for this problem, let's draw a number line. So here we have to this is going to be an open circle because it's less than it's not less than or equal to. So it doesn't include two and an open circle on this side. And this is everywhere that X can be X Is less than two is less than six. Okay, so now, which other statements are true and which aren't necessarily true? Okay, let's start with a zero is less than X Is less than four. So based on this concept of X being stuck between two and 6, can this be true? Can XB one here, X can be won in part a but it can't be true here in our given expression, So this is not true. Yeah, not necessarily true. Okay, how about E? Yeah, X zero is less than x -2 is less than four. Yeah. Okay. So what does this mean? Let's solve to isolate this X. What that means is just adding to to each part of this inequality to get X by itself And we get to is less than X is less than six. This is true. This is exactly what our expression says. It's exactly what it says two is less than X is less than six, so this is necessarily true. Okay, part C one is less than X over two Is less than three. What do we do with this one? This one we're going to isolate that X. Again. So here we do that by multiplying every part of the inequality By 2 to get rid of that to here. So once we do that two times one is 22 times X over two is just X. Two times three is six. This is true. This relates back to our given expression of two is less than X is less than six. So that is true. Okay, now part D. Mhm. 1/6 is less Than one over X Is less than 1/2. Okay. This one's kind of annoying, isn't it? Mhm. What can we do here? Well, one thing that we can do is try to clear this these denominators. Bye. Multiplying everything by 12 X. Let's just see how that works out. So 12 X times 12 X over six is too 12 x over X. is 12 12 X. Two X. Sorry, 12 X over six is two X. One X. 12 X over X. Is 12 and 12 X over two is six X. Okay. What what does that mean? It's divided by these tools to see if we can get this any clearer, This means X. It's greater than three is greater than two x. What what the heck does this mean? What does it mean? So it means that X is less than three and X. Is less than two X. If we think about it that way X is less than three which is less than two X. So let's look back at this and let's just plug in a number for X to see if it's true. Something that fits here. So like the number three Is three, less than three. Less than six. Well, no, not this isn't true. And the main reason that this isn't true is because three is not less than three. So if there was a less than or equal to sign this would work fine but there isn't one. So this is not true. Well what about if I plug into two is less than three is less than four that works but for any number Biggest than three, it doesn't work. And also if we just look at this, the only X has to be between two and six. Right, And three would be right here. So there's going to be very very little overlap between this and this so that is not true. Okay, Part E one is less than six over X is less than three for this. Let's try that method of plugging in the lowest so X equals to remember it can't actually be equal to to all I'm doing is testing the waters here to see if it works. We'll get one is less than 6/2. 3. Less than three. That doesn't work. And then for this one will get one is less than one is less than three. That also doesn't work. This is not true because three is not less than 31 is not less than one. Okay. Part F. Okay. Okay, Absolute value of X -4 Is less than two. What does this mean absolute value means whatever is out of here has to come out to be a positive. So one thing that we can do is split it into its negative and positive pieces. What? Because what it's saying is that The distance from zero is less than two. So this means s is X is less than six. If I take the positive side of this X minus four but there could also be a negative version. Right? This negative X minus four and if we plug that into the absolute value will get something positive but this negative thing can still exists on it's still exist on its own. Okay, so negative X is less than negative two. If I saw this X is greater than two, this is true, this is true. Okay, now just keep going. G negative six Is less than negative, X is less than -2. Okay, this one is a little tricky because remember with inequalities whenever you multiply or divide by a negative you have to flip all the inequality so that's going to turn into this, I multiplied by negative one, that is going to turn into that two. So this is true, this is exactly what we have up here, right? It's just flipped a little bit so if I switch it around you'll see it's exactly the same as though the one we were given to begin with. Finally E. F. G. H. H. Negative six. Mhm. Is oh I made a little mistake here actually. This one should be H. This one is age. So I'm doing this out of order because I copied it down wrong accidentally. This one is going to be G. Is less than two. The difference here is the negative which I missed the first time I read this over. So sorry about that. But let's just flip G. And H. We're going to do the same thing. We're going to see if this is true. six. We're gonna flip this sign this way. Going to flip this one this way. Okay. Is this true? No this is not true because remember negative X goes between two and six not negative two and six X cannot go that low according to our given expression here that we're governing all this by. So this is not true. In summary. A is not true. B is true, C is true. D. Is not true, he is not true. F is true. G is not true and each is true. Yeah.

Conjecture that if F of 000 and F prime of X exists at all, X values and f of X is less than or equal to exit, all points than F prime of X must be less than or equal to one. And let's craft this out. This isn't as intimidating as it might seem, so let's consider this point right here. F of zero must be zero. So we have to have a point for ffx right at the origin. We also know that f of X has to be less than or equal to X at all times. So we can't ever go past this dashed line. All of our functions have to stay on this bottom part of the graph. So these are all the point that I'm shading in right now. Those are all of the points where F of X is less enacts. A point up here would have a negative X value of positive y value, and that means that f of X y value would be greater than X. So Ah, this point right here just means that we can't have any jumped as communities or costs or anything like that. So let's think. Is there anything that we could dio that would stay in the shaded area? That would also have a positive slope greater than one? Well, we could do a straight line. Um, that, like, runs through here. I'll do it a different color that runs through here like this. But that doesn't work. It doesn't run through. Ethics here is equal to zero and still has a slope off one. So we've done nothing. Um, maybe we could have, like, a graph, you know, come up through here and just touch that point. And then maybe kind of, like, level out. But then it seems that there would be a uh huh. No, the graph wouldn't continue to the negative X values. It would just have vertical assam toe. So that doesn't That doesn't work. So it looks like this conjecture is true, because there's no way we could get an increasing function with a slow greater with the one that would never, ever cross this barrier right here in the middle

We can see from the graph that the value off effects is greater than zero when X lies between minus two and zero. From here, the value of X belongs to open interval minus 2 to 0, and well off X is greater than three, that is, ex belongs to open interval three to plus infinity. Now we need to find the points where affects has a negative value. Now from the ground, we can see that when X is less than minus two and actualize between zero and three that when acts belongs to upon interval minus infinity to minus two and acts belongs to open intervals 0 to 3, then the effects has negative value. Now we need to find the points where the value off FX is less technical. Do zero now. From the graph, we can see that when X belongs to or pond minus infinity to close minus two and when X belongs to close intervals 0 to 3, the value off effects is less than equal to zero. Now we need to find the points from the graph where the value of effects is greater than equal to zero. Now we can see from the graph that when X belongs to close interval minus 2 to 0 and when X belongs to close three to plus open infinity that the valley off effects is greater than or equal to zero. This is our answer.


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