Calling all math nerds! Need Help ASAP!

[fletch to 99]

Hey!

 

So I was wondering how to calculate turning points mathmatically between 2 roots if one of them is a double root. The equation is (-2x⁴+8x²)/(x³-3) or in factored form its: ((x-2)(x+2)(-2x²))/(x³-3)

 

This is what my graphing calculator produces: 

 

What I remember learning in math class was it was the middle of the 2 roots so in my head I'm thinking it should be x=-1, but on the graph its like x=-1.19 something rather. Could it be because x=0 is a double root?

[T.Vengeance]

The turning points are where the derivative of your function = 0 and it changes sign.

[WorldDominator]

The turning points are where the dertivative of your function = 0 and it changes sign.

Also undefined

[whatthe_fuzz]

Take the derivative of the ORIGINAL function and set it to 0. 

 

That will give you the critical value.

[fletch to 99]

The turning points are where the derivative of your function = 0 and it changes sign.

 

 

Also undefined

 

How do I do this mathmatically? Sorry. What I was taught in advanced functions was take the 2 x intercepts and divide them by 2 and that is the turning point.

[T.Vengeance]

How do I do this mathmatically? Sorry. What I was taught in advanced functions was take the 2 x intercepts and divide them by 2 and that is the turning point.

In 12 Calc, you'll learn that the derivative function, is the function of the slope of the tangent to your original function. In this situation, you can use the power rule and quotient rule to determine the derivative, then calculate what x value makes it zero.

[WorldDominator]

How do I do this mathmatically? Sorry. What I was taught in advanced functions was take the 2 x intercepts and divide them by 2 and that is the turning point.

Oh you aren't in calculus...I can't remember low  level math

[fletch to 99]

In 12 Calc, you'll learn that the derivative function, is the function of the slope of the tangent to your original function. In this situation, you can use the power rule and quotient rule to determine the derivative, then calculate what x value makes it zero.

 

I'm still confused as to how I would go about doing this. What is the power/quotient rule? I'm taking calculus next semester so I have no clue what a derivative function is

 

Edit: Could you maybe find a tutorial for this on YT for me? I've been looking but I'm unsure what to search for.

[fletch to 99]

Oh you aren't in calculus...I can't remember low  level math

 

Calc next semester

[T.Vengeance]

I'm still confused as to how I would go about doing this. What is the power/quotient rule? I'm taking calculus next semester so I have no clue what a derivative function is

Well, if you wanna get ahead, here's quick crash course. Like I said, a derivative calculates the slope of your tangent for a particular x-value of your function. If you take a simple function, like f(x) = x² you'd be using power rule to get the derivative function. Power rule states that you drop the exponent to multiply the function by it, then take that exponent and subtract 1. So the derivative, f ' (x) = 2x¹ or just 2x. That means that at x = 1, the slope of the tangent to x² = 2. So from this you can calculate when the turning points are since they happen when the slope of the tangent, and therefore the derivative = 0 and also changes sign. Changing signs means the slope goes from postivie to negative, or negative to positive. 

 

The quotient rule gets a bit more complicated. It states that if you have f(x)/g(x), to find the derivative, you take the derivative of the numerator multiplied by the denominator, subtracted by the derivative of the denominator multiplied by the numerator, and all of that divided by the denominator squared.

[fletch to 99]

Well, if you wanna get ahead, here's quick crash course. Like I said, a derivative calculates the slope of your tangent for a particular x-value of your function. If you take a simple function, like f(x) = x² you'd be using power rule to get the derivative function. Power rule states that you drop the exponent to multiply the function by it, then take that exponent and subtract 1. So the derivative, f ' (x) = 2x¹ or just 2x. That means that at x = 1, the slope of the tangent to x² = 2. So from this you can calculate when the turning points are since they happen when the slope of the tangent, and therefore the derivative = 0 and also changes sign. Changing signs means the slope goes from postivie to negative, or negative to positive. 

 

The quotient rule gets a bit more complicated. It states that if you have f(x)/g(x), to find the derivative, you take the derivative of the numerator multiplied by the denominator, subtracted by the derivative of the denominator multiplied by the numerator, and all of that divided by the denominator squared.

Ok I'll give that a shot but can you do me a favor and do it for my equation as well and give me the answer so I can verify/see where I went wrong? 

 

Edit: sorry to ask for so much but its for a math project that is due on Monday and for me to change the equation I want to use I would have to contact my teacher... which I can't do before monday. The turning points are critical because we need them to calculate the increasing and decreasing intervals of the equation.

[T.Vengeance]

Ok I'll give that a shot but can you do me a favor and do it for my equation as well and give me the answer so I can verify/see where I went wrong? 

Lol it's an ugly derivative so I'm not gonna do it . BUT, I'll give you a graphical representation of what it looks like.

The derivative is in black and the original function is in red. Notice how the derivative = 0 and changes sign (positive to negative) at the turning point.

 

Here's a graphing program I'd recommend. I use it all the time and it's great! http://www.padowan.dk/download/

 

In it's pure form, using the quotient rule, the derivative of f(x) is:

f'(x) = [(-8x³+16x)(x³-3)-(3x²)(-2x⁴+8x²)]/(x³-3)²

[TEDWARDS]

The answer is 3

[fletch to 99]

Lol it's an ugly derivative so I'm not gonna do it . BUT, I'll give you a graphical representation of what it looks like.

The derivative is in black and the original function is in red. Notice how the derivative = 0 and changes sign (positive to negative) at the turning point.

 

Here's a graphing program I'd recommend. I use it all the time and it's great! http://www.padowan.dk/download/

 

In it's pure form, using the quotient rule, the derivative of f(x) is:

f'(x) = [(-8x³+16x)(x³-3)-(3x²)(-2x⁴+8x²)]/(x³-3)²

So if I follow this video using the derivative you gave me it should all work out? http://www.youtube.com/watch?v=YGcLdzQeEck

 

Basically I have to simplify the derivative and solve for 0?

[KaareKanin]

d/dx(-2x4+8x2)/(x3-3)=-((2x(x^5+4x^3-12x^2+24))/((x^3-3)^2))
 
Edit:
Dammit T.Vengeance, you editing your posts?

 

Edit 2:

fletch to 99, you have a local maximum at x=0 (i.e. the top of the graph where it turns down again), local minimum at about x=-1.15067 (bottom of graph, where it turns up) and a discontinuity at about x=1.44224 

[T.Vengeance]

So if I follow this video using the derivative you gave me it should all work out? http://www.youtube.com/watch?v=YGcLdzQeEck

 

Basically I have to simplify the derivative and solve for 0?

Yeah basically since it's a fraction, of course it = 0 when the numerator = 0. Factoring & simplifying, it = 0 when x = 0, and some other number lol.

[T.Vengeance]

Edit:

Dammit T.Vengeance, you editing your posts?

Lol I'm the master of math mistakes....So when I post, I'm pretty sure I'm right...then it catch one. And then I'm sad.

[fletch to 99]

Yeah basically since it's a fraction, of course it = 0 when the numerator = 0. Factoring & simplifying, it = 0 when x = 0, and some other number lol.

So I followed that video and got to 0=-8x⁹+6x⁸-24x⁴+40x³-54x but I have no clue how to solve that. Normally we would use the quadratic formula or long division but neither work here.

[Sarcasm]

So I followed that video and got to 0=-8x⁹+6x⁸-24x⁴+40x³-54x but I have no clue how to solve that. Normally we would use the quadratic formula or long division but neither work here.

If that's the derivative you got of the original function, I'm pretty sure you did something wrong...

[T.Vengeance]

So I followed that video and got to 0=-8x⁹+6x⁸-24x⁴+40x³-54x but I have no clue how to solve that. Normally we would use the quadratic formula or long division but neither work here.

It's beyond my level of math TBH. Learning 12calc&vectors is more on the concepts, less on the actual solving part. 

[Sarcasm]

It's beyond my level of math TBH. Learning 12calc&vectors is more on the concepts, less on the actual solving part. 

You're using that turd of a Nelson book aren't you?

[T.Vengeance]

You're using that turd of a Nelson book aren't you?

Nah, even more turd. Dat mcgrawhill one. 

[Sarcasm]

Nah, even more turd. Dat mcgrawhill one. 

Which one? I remember Ontario standard being the green Nelson book. This one

[T.Vengeance]

Which one? I remember Ontario standard being the green Nelson book. This one

 

I'm guessing you were in school when they still had that OAC credit? I'm not too sure when Ontario started using this, but the books in my school were pretty old.

[Sarcasm]

If that's the case, I recommend "obtaining" a digital copy of the Stewart book, and using that to teach you the theory. It's a hella lot better than either of these textbooks.