# Rational Function Application

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I know how to solve it. You can use synthetic divsion to figure out the equation that the tree gave you: $x^3=3x+2$

Therefore: $x^3-3x-2=0$. I strongly recommend you to use synthetic division.

Thanks a lot! I think GOT IT!!!!! $-20x(x-2)(x+1)^2=0$. So i found the zeros to be x=0,2,-1....i found there to be a local minimum at x=0 and a local maximum at x=2.. can someone please explain the significance of the second derivative in regards to interpreting the shape of the function graphed? I just dont really know why it is necessary to find it and when to....should i forthis function? Thanks

............I am so lost now. I graphed the function to find only one turning point

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Ya it makes sense but if i cant explain how to get it then i cant use it
Guessing is a perfectly valid method of working something out, so is one line of intuition so long as you can show that it is correct.

I don't know how to solve cubic equations but that doesn't mean I can't give a correct answer for it.

Think back to when you were first introduced to quadratics, you would be given something like x2+9x+14 and when asked to factorise that you would just give (x+2)(x+7), but you wouldn't show the steps in between.

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Guessing is a perfectly valid method of working something out, so is one line of intuition so long as you can show that it is correct.

I don't know how to solve cubic equations but that doesn't mean I can't give a correct answer for it.

Think back to when you were first introduced to quadratics, you would be given something like x2+9x+14 and when asked to factorise that you would just give (x+2)(x+7), but you wouldn't show the steps in between.

ya i did that in my post above yours..i got x=0,2,-1

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From $x^3-3x-2$, I used synthetic division, and I got $(x-2)(x+1)^2$, therefore the critical points are: $2,-1$.

I don't really understand why you had to find the second derivative of the function. Just input 2 and -1, in the original function, and you'll get the local minimum/maximum.

Note: If you want to know if the shape is concave in positive or negative direction, then you use test points between the critical points. Input the test points in the first derivative of the function, and you'll find the shape of the original function.

If you don't understand this, then I can give you some examples.

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From $x^3-3x-2$, I used synthetic division, and I got $(x-2)(x+1)^2$, therefore the critical points are: $2,-1$.

I don't really understand why you had to find the second derivative of the function. Just input 2 and -1, in the original function, and you'll get the local minimum/maximum.

Note: If you want to know if the shape is concave in positive or negative direction, then you use test points between the critical points. Input the test points in the first derivative of the function, and you'll find the shape of the original function.

If you don't understand this, then I can give you some examples.

The thing is that i graphed it and there is only one local extrema at x=2 which is a local maximum (2,5)...so what is the point of finding the second derivative...my teacher said that the first derivative finds the local extrema and the second derivative will find the concavity...i'm confused

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Yes, you're right. I did some research on "concavity", and it said that you have to find the second derivative of the function to find the point of inflection.

My calculation of the second derivative is different than yours. Since $f(x)=-20x^4+60x^2+40x$ and $f(g)=(x^3+x^2)^2$, you use the quotient rule.

I found the calculation of the second derivative to be: $\frac{240x^{10}+160x^9-880x^8-510x^7-200x^6+200x^5+40x^4}{(x^3+x^2)^4}$.

Such a painstaking calculation!! So you put $f''(x)=0$, and you should be able to find the value of x.

P.S. Is there any derivative tool on the internet so we can check if it's right?

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Yes, you're right. I did some research on "concavity", and it said that you have to find the second derivative of the function to find the point of inflection.

My calculation of the second derivative is different than yours. Since $f(x)=-20x^4+60x^2+40x$ and $f(g)=(x^3+x^2)^2$, you use the quotient rule.

I found the calculation of the second derivative to be: $\frac{240x^{10}+160x^9-880x^8-510x^7-200x^6+200x^5+40x^4}{(x^3+x^2)^4}$.

Such a painstaking calculation!! So you put $f''(x)=0$, and you should be able to find the value of x.

P.S. Is there any derivative tool on the internet so we can check if it's right?

I wish i knew of one and yes finding the second derivative is a pain because one simple error could completely ruin your anwser and it takes forever to find...

I got $\frac{40x^{9}+40x^8-240x^7-560x^6-440x^5-120x^4}{(x^3+x^2)^4}$....how am i to solvew this? Synthetic Division would take forever.. Damn...by the end when i finally bestow my teacher with this, i am going to have an enormous ulcer.

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Did you use the Chain Rule?

You know that the Quotient Rule is $\frac{f'g-fg'}{g^2}$.

On the $fg'$ part, you have to use Chain Rule to find the derivative of $g$. ($2(x^3+x^2)\cdot3x^2\cdot2x$)

If you don't know how to do it, then it's out of your league. I didn't learn Chain Rule in high school myself. This is more of an university expertise. You can just tell your teacher it's an unacceptable question, or you could just leave it alone.

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Did you use the Chain Rule?

You know that the Quotient Rule is $\frac{f'g-fg'}{g^2}$.

On the $fg'$ part, you have to use Chain Rule to find the derivative of $g$. ($2(x^3+x^2)\cdot3x^2\cdot2x$)

If you don't know how to do it, then it's out of your league. I didn't learn Chain Rule in high school myself. This is more of an university expertise. You can just tell your teacher it's an unacceptable question, or you could just leave it alone.

No i dont know of the chain rule...i was taught the quotient rule 2 weeks ago...is it difficult? I'm going to confront my teacher because function is unlike anything we have done. I asked a friend he never even knew how to get the first derivative solved too.

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This is what your originial function and the first derivative should look like:

You should be able to notice the change of slope at x=2.

The second derivative function shape is kinda very wierd, so I'll try it one more time. I can post the graph of the second derivative if you are interested.

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This is what your originial function and the first derivative should look like:

You should be able to notice the change of slope at x=2.

The second derivative function shape is kinda very wierd, so I'll try it one more time. I can post the graph of the second derivative if you are interested.

naw its quite alright i got a graphing program unless it is benefiicial to understand why the shape is weird...so i know that at x=2 there is a turning point but then what becomes of the zeros that i found to be x=0 and x=-1...becasue by the look of the graph nothing significant occurs at those 2 zeros. thanks you are very helpful.

Anytime.

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Anytime.

I wish i could just understand this...i wish i could just get math like i use to this is really depressing me...maybe i'm just good for nothing and i show just stick with what im good at and just go to college instead of university...Thanks everyone for trying to help someone brain dead like me.

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You know what's funny? Last year in high school, my math teacher refused to teach me the Chain Rule. So I had to do the research myself, and I couldn't understand it, so I got really depressed. ender7x77, just get some sleep, and you'll feel better tomorrow.

In university, (actually around a month ago), I finally learned it, and I was really happy. It just takes time before you can understand - requires a lot of practices, that's all. Don't worry, you're fine.

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You know what's funny? Last year in high school, my math teacher refused to teach me the Chain Rule. So I had to do the research myself, and I couldn't understand it, so I got really depressed. ender7x77, just get some sleep, and you'll feel better tomorrow.

In university, (actually around a month ago), I finally learned it, and I was really happy. It just takes time before you can understand - requires a lot of practices, that's all. Don't worry, you're fine.

Thanks for the encouragement the only problem is that this is an assignment question and its due tomorrow and i have no idea how to do it...Once again i thank u though!!! I guess i'll ask my teacher before class begins

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Im sorry but I dont understand what exactly is hard hard about the chain rule?..Its the easiest to remember and apply, and most people have learnt it 2 years before the end of high school in australia, which isnt particularly smart..

we are thinking of the same rule, arent we? dy/dx = dy/du *du/dx ???

its basically multiplying the rates of change to get a total rate of change. eg Im going 2 times as fast as you. A car goes twice as fast as me. Multiply and you get the car, goes 4 times as fast as you. SImple.

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well i asked her but by that time it was to be handed in and guess what...the question is much simpler than it appears. For those interested, you have to reduce it where you will get a much simpler f to solve and a removable discontinuity of x=-1....i am so upset... Thanks everyone for trying to help me out. It mean't a lot.

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By the way, the zeros x = 0 and x = -1 are vertical asymptotes.

EDIT: oops didnt see page 3 of this thread. This imformation is a tad late

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