# Re: minimising

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Posted by T.Gracken on August 26, 2002 at 13:08:10:

In Reply to: Re: minimising posted by KK on August 25, 2002 at 18:57:24:

: How about this one? y=(2000+20x+10x^1/2)/x
: I got y'=(-2000x^-2) - (5x^-3/2)
: Why is it a square root can't have negative? How come this one does not have a minimum?

It's not that you can't have the square root of a negative number. It's that if you want to work with real numbers (numbers that can be written as decimal numbers), you can't have a square root of a negatvie number.

Why??? besides zero, any decimal number multiplied to itself is positive.

The reason there is no minimum in your new problem, is that 0 (zero) is no longer in the domain. Even though the function is still increasing on the interval (0,inf), but since 0 is excluded from the domain, there is no minimum.

remember that your first problem was defined on a half-open interval. You may want to look at your text under Rolle's theorem & the Mean value teorem for more insight to this type of question.

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