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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: madoscar65 on June 22, 2011, 07:06:09 pm

Title: Local minimum question
Post by: madoscar65 on June 22, 2011, 07:06:09 pm

Hi Vners,
Just been stuck on proving this: What values of n for graph of y=x^n to have a local minimum? I know that an even number will have a local minimum but I don't know how to prove it. Any help will be appreciated.

Thanks :)

Edit: Problem solved :)

Title: Re: Local minimum question
Post by: taiga on June 22, 2011, 10:18:30 pm

Spent a ridiculous amount of time thinking about this, such a good question!

Glad you've got it, I'll post the solution if anyone wants it, otherwise ceebs :D

Title: Re: Local minimum question
Post by: epinephrine on June 22, 2011, 10:37:53 pm

Yes I would like to see the solution Taiga  :D

Title: Re: Local minimum question
Post by: taiga on June 22, 2011, 11:34:37 pm

y = x^n

Assume n is positive.
If n=0 it's just a straight line, for 0<n<1, there are no stationary points, hence there can be no minimum.

dy/dx = n*x^(n-1)

If n is odd, for example 3

dy/dx = n*x^2.
Given n is positive, there is no way here the gradient can be negative due to the squared number.

To have a local minimum, the gradient needs to be negative at some stage then become positive. For any odd number the gradient clearly can not be negative and therefore no local minimum can exist for any odd number.

That's the furthest I can explain it within the scope of maths methods.

EDIT: asked OP and he had a different solution which sounds very reasonable

Title: Re: Local minimum question
Post by: madoscar65 on June 23, 2011, 09:50:28 am

Edit: misread

Title: Re: Local minimum question
Post by: taiga on June 23, 2011, 10:37:06 am

What do you mean?

I thought you had your own set of solutions that you might have wanted to post (OP in this case means original poster)

Title: Re: Local minimum question
Post by: madoscar65 on June 23, 2011, 03:40:04 pm

LOL sorry my bad, misinterpretation :(