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## Homework Statement

Show that the maximum and minimum values of y occurs when x^3=8+- 2(14)^1/2

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- #1

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Show that the maximum and minimum values of y occurs when x^3=8+- 2(14)^1/2

- #2

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Plus check this: x^3=8+- 2(14)^1/2

is it : ##x^3 = 8+(-2)(14)^{1/2}## ??

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What is y defined to be?

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-x^2)

- #5

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You should have definition for y, right? Else what would you differentiate?

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Please post the full question.

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Y^3=6xy-x^3-1

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- #9

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Okay,

##y^3=6xy-x^3-1##

and

##x^3=8\pm2\sqrt14##

right?

Now how have you decided to proceed?

##y^3=6xy-x^3-1##

and

##x^3=8\pm2\sqrt14##

right?

Now how have you decided to proceed?

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Yup correct

- #11

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I think I have posted the attachment now,sorry I am a first timer here

Well, that completely makes sense now. What have you attempted?

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- #13

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Right.As I know at stationary point,dy/dx=0 so from dy/dx=(2y-x^2)/(y^2-2x),from there I can find y=(x^2)/2

Ok that will fetch you the maximum and minimum of y.then I substitute x=2+2(14)^(1/2) and x=2-2(14)^(1/2) to get y values right?

I'm not sure if I understood this one.Then I reciprocate the dy/dx =dx/dy and then I do the sign test? But I got all + signs which means I fail to prove y has maximum and minimum values.

But as soon as you get the ##y=x^2/2##, can you use it and the 'given' to solve for values of x?

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- #15

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Right.

Ok that will fetch you the maximum and minimum of y.

I'm not sure if I understood this one.

But as soon as you get the ##y=x^2/2##, can you use it and the 'given' to solve for values of x?

I haven't attempt which I didn't thought of it that but as I sub it back into y^3, I get x=2 and y=2,so now I have 3 values of x and I can find 3 values of y,and then I need to sub into the second derivative test or using a sign test but If I do a sign test there,there are two variables,how to do it ? @@

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You mean you substituted y=x^2/2 in the original y equation. What did it yield? Which are the three x's and y's?but as I sub it back into y^3

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- #20

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If typing in was difficult, a snapshot of your work will do.

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I have used the calculator to change the x^3 values into 15.84 and -5.84,but when I substitute the x values and y values (how I find y values,I get it from y=(x^2)/2,substituting the x values as stated above) into my second derivative but in the end I got all positive value which means only local minimum values only. So I did it wrongly?@@

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- #23

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How did you get this? Probably by the previous x^3 equation which they have given?I have used the calculator to change the x^3 values into 15.84 and -5.84

but when I substitute the x values and y values into my second derivative but in the end I got all positive value which means only local minimum values only.

The implication is right, but I don't know if all are minima or maxima or both, because I haven't worked it out. [STRIKE]If all are minima, that doesn't have anything to do with the answer they expect.[/STRIKE] Okay I saw what the question asks for.

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- #26

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Oooo

- #27

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Okay I see..

Go step by step: You found first derivative and equated it to zero. Implies y=x^2/2. Next step is to use this and solve for 'x'. See if you can do this by substituting in the givens.

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I don't think I know how do I get x^3=8+-2(14)^{1/2} from the equations provided

- #29

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I don't think I know how do I get x^3=8+-2(14)^{1/2} from the equations provided

I know, it

>You used a calculator to approximate the given x value ( I think this is unfair).

>Both x values are yielding minima (right?)

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- #31

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- #33

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can I use the sign test to find the maximum and minimum values from this kind of equation?

You mean the second derivative test? It says only if the point is maxima or minima. It doesn't give you the value.

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- #35

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Oh that one, for that you need to know the interval, and it requires you to solve for x.Sign test is a test which you use a value which lies inside the interval of a variable,example x then substitute it into dy/dx to find the sign and we can find the maxima and minima.

but anyways I manage to prove it by using 2nd derivative but not sure correct or wrong.

Correct as long as you don't assume the values of x with a calculator and as long as y has both maxima and minima.

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