Re: continuous functions

magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin) writes in article <dltdpm$nt4$1@xxxxxxxxxxxxxxxxxx> dated Mon, 21 Nov 2005 21:21:58 +0000 (UTC):
>In article <27774017.1132607662058.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
>jaspal <jsdhariwal@xxxxxxxxxxx> wrote:
>>Show that the equation 4x^3 - 6x^2 + 1 = 0 has three real solutions.
>>Through factorization, I have obtained values of x= 1/2, (1+ or - sq. root of 3)/2.
>>Can someone please tell me how to proceed from here?
>
>What do you mean, proceed? You're done! You have found three
>solutions, all of them real.

In fact, he went beyond the call of duty and found out what they were!

I'm much lazier. I would have found the zeroes of the derivative and
evaluated the function at each of those points and at +-oo. Then argued
that the function must have a zero for each sign change due to its
continuity.

--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
.

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