Help plz
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@Narad @jhonyy9
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Is there a method to find the max range aka point of vertex?
Yes, it's called taking the Left Hand limit. You need to do: \[\lim_{x \rightarrow 1^-} \frac{ 2 }{ x^2 + 1}\]
Then take the right hand, and see if they are continuous (which they are based on the graph). Then you can say the max/min. gtg
I’m literally on page 2 bro. This ain’t the season for limits just yet
i can take the derivative of the denominator tho....
I need a power nap brb
Yes, there is a method to find the range.
Alright ear, how do we proceed to find the vertex? I believe there’s a specific equation/formula u use right?
You said you wanted to find the range of the function. I can help you find that.
Plz guide me
Okay, so the first step is to replace \(f(x)\) with \(y\): \(y = \dfrac{2}{x^2 + 1}\)
Okie dokie but let me do it
Next, multiply both sides by \(x^2 + 1\). Of course you're going to do it
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What happened to the \(y\)?
You need that to find the range.
I replaced the y jus like u said bro
Isolate and solve?
I said to replace \(f(x)\) with \(y\)
I did not say to remove the \(y\)
But u said multiple both sides by x^2+1 tho
Let’s try this again
Yes, I did. Multiply both sides of \(y = \dfrac{2}{x^2 + 1}\) by \(x^2 + 1\) . Does doing that remove the \(y\)
I'll give you another chance to do it correctly.
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You must use parentheses to indicate that \(x^2 + 1\) is a multiple of \(y\)
\(y(x^2 + 1) = 2\)
Wait, I needa buffer of 10 chances before u take over K? 2 ain’t enough. I wanna do the heavy lifting in these questions. Plz jus gimme hints in the process 😇❤️
Okie dokie. Let me fix that
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I don't really have time to give you ten chances.
But plz tho *~*
Next apply the distributive property. Distribute the \(y\) over \(x^2 + 1\)
Okie dokie
And simplify the Right Hand Side
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Very good. Next, subtract 2 from both sides.
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Okay, now write the equation in the form \(ax^2 + bx + c = 0\)
Isn’t it already in that form tho?
I need a hint plz
Basically, your hint in this case is that the \(y\) in the first term acts as a coefficient in this case.
It is important to write the equation in standard form or else the rest of the steps will be too confusing for you.
Thinking...
The most I can think of is making the into pairs like we do when we factor....
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YES! Very good.
Smh....😐😡
U could have said to *rearrange* them into standard form ....lol
Now, what we will do next is find the expression for the discriminant. The discriminant is \(b^2 - 4ac\). So we must identify our \(a,b\) and \(c\) values. What are they in this case. Remember that our quadratic expression is of the form \(ax^2 + bx + c\)
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I knew you would guess that, however, if you look closely, you'll notice that we don't have a \(bx\) term.
What we have is \(ax^2\) and \(c\) terms but no \(bx\) term. Let me show you:
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Let me SHOW you
Wait, so do we set the whole term to solve for BMX?
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You have to identify what \(c\) is first before you can use that
Honestly there has to be an easier way to solve this...
I told you this would get confusing
Oh there is. But this is the easiest algebraic way
Bet this can open a wormhole straight into 9th dimension...smh
So now that you know a b and c, use the discriminant to find the expression. Let me clarify the a, b, and c again so that you don't confuse yourself again
Okie dokie
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Eye sea
I need ur help. Idk what to do from here...the most I can think of is factoring
And that’d make everything even more messier
The goal was to apply the discriminant \(b^2 - 4ac\) (I mentioned this earlier)
I know this is a lot to process for you but this is a simple method. You're just not used to it
Wait..|dw:1560641836395:dw|
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The discriminant shouldn't contain any \(x's\) just the coefficients. I circled them for you remember?
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You're making it more difficult than it really is unfortunately
My bad... |dw:1560642002465:dw|
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The discriminant is solely \(b^2 - 4ac\) you don't need to add the "equal zero" part to it because we are going to apply something else here.
I was jus making it ultra clear
No, the problem is, it is not equal to zero. That's not what we apply here.
I jus wanted to find the range tho....*~*
This is too tedious
We ARE finding the range.
Only because you're not used to it. You're "guessing" and making assumptions and not asking enough questions before doing the algebra.
In order to generate a real number value the discriminant is always going to be GREATER THAN or equal to zero.
K fine. |dw:1560642341893:dw| What do i do now?
Well, now you've just abandoned the discriminant expression and went back to the quadratic equation. Do you remember the discriminant expression we just found?
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Yes, and what was the expression we generated after applying the a, b, and c values?
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Yes, you did that already here:|dw:1560642528024:dw| we just need to make that GREATER than or equal to zero.
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The discriminant \(b^2 - 4ac\) is a FORMULA that you apply when you have a quadratic expression \(ax^2 + bx + c\)
And you know the \(a\), \(b\) and \(c\) values
Help me finish this question before I lose my sanity....
But yes you have it now
What do i do next?
For the most part you have it
Wait? That it?
No, that's not what I meant when I said that. We still have work to do.
.....
I'll explain what I meant when we finish
Okie dokie. What’s next?
The next step is to add \(-4y(y - 2)\) to both sides of the inequality.
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Sorry
See this is why I'm allowing you do this. It's to expose your weakness. I said to add \(-4y(y - 2)\) to both sides of the inequality. Here's how to do that: |dw:1560643037063:dw|
That’s what i was thinking but was kinda afraid
So in other words, what we have now is \(4y(y - 2) \le 0\)
The next step is to divide both sides by 4
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Actually, I see what you did. You had it right. You're just re-posting steps unnecessarily
We flipped the inequality by slipping the signs ..I get it
Flipping
No, you had it right the first time. That's my bad. SMH. You don't need to re-post steps.
Please, ignore what I said
It seems we cannot do this without confusing you
I wanna cry blood ..... whole dawn one sec
But we're almost done. Let's not give up hope yet
What we have now is simply \(y(y - 2) \le 0\) which is something I'm sure you've seen before.
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See when you re-post steps we've already gone over, that is what confuses me. May I ask why you're re-posting previous steps? I work in a progressive manner which is to do one step then go to another without re-posting previous steps unnecessarily.
Cuz I wanna see where I made a mistake...
The work you posted is correct though. I assume you're re-posting to make sure you understand.
What you re-posted is correct. You did not make a mistake in doing it.
Okie dokie, can we plz finish this ?
Yes, continue with dividing both sides by 4
What’s next?
Well, you stopped at a step we already did
But please do not "flip anything". That's not what we're doing here
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YES. Correct
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You're introducing confusion into the solution. What are you trying to understand here? The discriminant is \(b^2 - 4ac\) we don't use \(x\) in the discriminant
The goal is to find the range of \(y\) Forget about \(x\) at this point
okie dokie... ignore what i said earlier... sorry
Thank you
So as I was saying before the confusion, what we have is \(y(y - 2) \le 0\). But we must know how to proceed from here to find the solution.
so what do we do?
The next step is to find the critical points. In other words we have to apply the zero product property: If \(ab = 0\) then \(a = 0\) or \(b = 0\)
In other words, we set the factors of the discriminant expression equal to zero then solve for \(y\)
okie dokie..one sec
Believe me when I say all of this is easy, you've just never worked with any of this before and it seems tedious. It is not.
It can be done in less than two minutes with familarity
It's only confusing if you don't know what to do with the \(x's\) and \(y's\) and if you're still trying to catch up on your algebra
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I’m wrong;/
Let me re-post: The next step is to find the critical points. In other words we have to apply the ZERO PRODUCT PROPERTY: If \(ab = 0\) then \(a = 0\) or \(b = 0\) In other words, we set the factors of the discriminant expression EQUAL TO ZERO then solve for \(y\)
There are some key symbols to apply there
I'll give you one more chance. Then we have to move on.
I give up...tell me what to do
Hint: Identify the factors.
What are the factors in this case?
There are two factors: (second hint)
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Okay, at this point, I'm going to think you have a reading comprehension problem. I'm going to post what I wrote one more time. And this time you are to read it aloud before doing anything
The next step is to find the critical values In other words we have to apply the ZERO PRODUCT PROPERTY: If \(ab = 0\) then \(a = 0\) or \(b = 0\) In other words, we set the factors of the discriminant expression EQUAL TO ZERO then solve for \(y\)
The key word in this case is EQUAL
I’m going to read this out loud....
not greater than or less than
Also I asked you to identify what the two factors are first before doing anything
Omg...wait
Identify the two factors first
Let me know what they are
Finish this sentence: The two factors are ... Blank AND blank. Fill in the blank
We're so close to being done
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The word *factor* is throwing me off. Is there another word to describe this?
Don't beat yourself up. You did the best you could. Two factors of an expression are two things being multiplied. In this case the two factors of the discriminant are \(y\) and \(y - 2\)
Those are the two factors we are to set equal to zero.
Factors have a different meaning in my head sooo that threw me off
It's okay. You know what they are now. Stress relieved.
U should’ve used a more formal aka none mathematical language
You mean INFORMAL (non-mathematical) language?
Yuhh that lol
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Correct
Gimme a second... I need to shove a flame thrower up in my Potato real quick....
That just shows that I could’ve solved this question much more easily had u used simple none mathematical terms
Now we have found the limits of the range. At this point though, we need to figure out if these values are "included" or "excluded".
Your fancy mathematical terms sparked the panic trigger inside me
Wait, we not done?
We have to take the \(y\) values we found and plug them in to the original equation. \(y = \dfrac{2}{x^2-1}\)
We're just about done
If you were more familiar, we'd already be done
Yo hero, I love u but this ain’t cup of my tea mate.
Bro, we don’t need limits to solve the range...
Forget the term "limits". I'm replacing it with "extremes"
Ur trynna solve this question as a PhD in mathematics mate
We found the extremes of the range. We just need to figure out which value is included and which value is excluded.
Nope, this is the non-calculus algebraic method of finding the range.
Not PhD level
It would take you two minutes or less to do the whole thing if you didn't need to brush up on algebra.
R u telling me that we need to test which values fall within the extreme and which don’t?
Everything I just did is High school Algebra level
What we need to do is figure out which value is included and which is excluded so that we express the range correctly.
bro I knew that the range of the function is 2...cuz any value of x doesn’t let y exceed beyond that point
Trust me, I wasn't even using formal mathematics here. I know that you knew, however, if on a test, they ask you to find the range of a function, this is the correct method to find that range.
The range is (0,2]
In other words zero is excluded and y is included
But to find that without using a calculator or a graphing tool, you use the method I just showed you.
okie dokie. i have more gutt wrenching questions coming up in a second... thanks for you hard work and time ....<3
All you need to do is realize that we did nothing but high school mathematics to find the range.
that's a fairly simple function i could've used a hand full of x values to realise the answer... but we'll see if we need to use your method in the upcoming questions
I did skip a step but I only avoided it because I fear that it might trigger more panic from you.
thanks mate :)
i need a water break...brb :)
we could actually do a live stream on twitch and make some money off of this...kinda LOL
that way, we'll have a chance to promote this site and run some ads to make some money lol js lol
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