- #26

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Oooo

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- Thread starter Liang Wei
- Start date

- #26

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Oooo

- #27

PhysicoRaj

Gold Member

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Yeah,the x values are from the x^3 that they provide,when you substitute those values,you will find that there are many zero values so it is able to get the d2y/dx2= +value which is greater than zero so both values of x are local minimums

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.

- #28

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I think they expect you to arrive at ##x^3=8+-2(14)^{1/2}## rather than assume it at the first. This is what is going wrong.

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

- #29

PhysicoRaj

Gold Member

- 479

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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?)

- #30

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- #31

PhysicoRaj

Gold Member

- 479

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- #32

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- #33

PhysicoRaj

Gold Member

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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.

- #34

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- #35

PhysicoRaj

Gold Member

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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 dont assume the values of x with a calculator and as long as y has both maxima and minima.

- #36

- 18

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Ok thanks

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