Re: [Math4u] Re: Maxima & Minima

I agree with the second Jim that f(2) exists and it is a minimum.

I agree with the original question that if a function is differentiable, there must be a minimum between 2 maxima.  But I think Jim FitzSimons is talented so maybe I am missing something.

Going back to Jim FitzSimons message

f(x)=-x*(x-4)*(x^2-4*x+6), x not equal 2

How did Jim come up with such an example, why is this factored and why did Jim say there are maxima at x=1 and x=3?

f(x)= -x^4 + 8x^3 – 22x^2 + 24x
f'(x) = -4x^3 + 24x^2 –44x + 24

If I substitute 1 or 3 for x into the above equation, I don't get zero so there is no maximum or minimum at x=1 or x=3

Regards, Brian Jensen

From: Jim

To: Math4u@yahoogroups.com

Sent: Sunday, September 23, 2007 11:43 AM

Subject: [Math4u] Re: Maxima & Minima

A different Jim disagreed with Jim FitzSimons:

>Unless I misread your function f(2)=8 so it does exist and is a minima.

--- In Math4u@yahoogroups.com, "w7anf" <cherry@...> wrote:
>
> False, here is an example.
> f(x)=-x*(x-4)*(x^2-4*x+6), x not equal 2
> f(x) has maximas at x=1 and x=3.
> The minima should be at x=2, but the function does not exist
> at x=2. f(x) is differentiable at x=2.
> Jim FitzSimons
>
>
> --- In Math4u@yahoogroups.com, "sanjivadayal" <sanjivadayal@> wrote:
> >
> > Whether the following statement is true or false?
> > "If f(x) is a differentiable function, then there must be at least
> one
> > minima between two maximas".

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