rational functions

[psi20]

f(x) = (x+3)(x-3)/(x+2)(x-2)

 

This yields 6 quadrants with a parabola-like curve in the middle.

 

The curve isn't a parabola, correct?

 

How do you locate the center of this curve?

 

f(x) = x/(x+2)(x-2)

 

The inner graph looks like x^3. Is its center the origin?

[Dave]

f(x) = (x+3)(x-3)/(x+2)(x-2)

 

This yields 6 quadrants with a parabola-like curve in the middle.

 

The curve isn't a parabola' date=' correct?[/quote']

 

Correct.

 

How do you locate the center of this curve?

 

f(x) = x/(x+2)(x-2)

 

The inner graph looks like x^3. Is its center the origin?

 

I don't know: never heard of the "centre" of a graph before.

[psi20]

Oh, I didn't know that not all rational functions were symmetric.

 

What I meant was how do you, if you can, use derivatives for rational functions to find out when the slope is 0? How do you find the vertex of the parabola-like shape?

[bloodhound]

i never heard of centre of a curve. it maybe possible to find the centre of curvature of the function at a given point

[Guest ajmmacleod]

If you just want to find the stationary point (The point where the gradient is zero) You have to diferentiate the function, and set it equal to zero. This works because when you differntiate, you are finding the gradient, so setting it equal to zero will find all the points along the graph where the gradient is zero.

 

Because you're finding all the points where dy/dx = 0, you'll end up with several answers, and you'll have to decide which x value best matches up with the stationary point that you're after, then you can simply sub this x ordinate back into the original y= equation, to find the y ordinate.

 

Hopefully I didn't lose you anywhere in all of that, If you have any problems give us a yell.

 

Cheers,

Adam

[Dave]

Just as an example: find the stationary points of [math]f(x) = x^2 + 2x + 5[/math].

 

[math]f'(x) = 2x + 2[/math].

 

Setting this to zero, we can find stationary points. So:

 

[math]f'(x) = 0 \Rightarrow 2x + 2 = 0 \Rightarrow x = -1[/math]

 

Sticking this back into our original equation, we get that [math]f(2) = 4[/math] so we have a stationary point at (-1, 4).

[psi20]

I can't figure out how you do it with rational functions like:

2(x^2 + 9)/(x^2 - 4)

[Guest ajmmacleod]

What part are you having trouble with? Differentiating, or the stuff following that?

 

To differentiate that equation you either have to do long division or use the quotient rule

 

let f(x) = 2(x^2 + 9)

let g(x) = x^2 - 4

 

 

Derive those 2 simple functions, and chuck everything into the rule.

 

Adam

[Dave]

Or if you're so inclined, you can even use the product rule. Just set the denominator to [math](x^2 - 4)^{-1}[/math].

[psi20]

Oh, so if you have f(x) = 2(x^2 + 9)

and g(x) = x^2 - 4,

f '(x) = 4x and g '(x) = 2x

 

When you set them both to 0, you get 0 = 4x and 0 =2x, so x=0.

Then you plug it all in and get f(0) = 4.5

 

Did I do it right?

 

f(x) = 2x/(3x^2 + 1)

Let a(x) = 2x and b(x) = 3x^2 + 1

a '(x) = 2, b '(x) = 6x

 

So set both to 0? 0 =2, 0 = 6x

x=0

So plug it back in and it says (0,0), but that's not the answer. There's something I'm missing here.

[Guest ajmmacleod]

you must plug the f(x) g(x) f`(x) and g`(x) equations into the quotient rule, and this will give you the derivative of you're original y= equation.

 

If you missed the link in my previous post, here it is again:: http://mathworld.wolfram.com/QuotientRule.html

 

Cheers,

Adam

[psi20]

Ah, thanks

[bloodhound]

I dont get what all these replies about derivatives has to do with the original question in the first place

[psi20]

You know in some rational functions there's parabola looking things within asymptotes, I wanted to find the point where the gradient or slope is zero.

[bloodhound]

oh ... you meant centre as in the midpoint. sorry, i mistook it for centre as in centre of a circle. never mind