Posted by T.Gracken on August 26, 2002 at 13:09:56:
In Reply to: Re: minimising posted by Soroban on August 26, 2002 at 12:38:16:
: : 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?
: Good questions!
: Hope I can clarify the mystery.
: The distinctions are quite subtle.
: If the question is: "x^2 = 9. What is x?",
: the answers are +3 and -3.
: If the question is: "What is sqrt(9)?"
: (with a radical over the 9), the answer is 3.
: What's the difference? In both cases we take
: "the square root of 9", don't we?
: In the first case, we are solving a Quadratic
: Equation, so we must give BOTH answers.
: In the second case, the square-root sign is
: GIVEN in the problem. And by "convention" (or
: "tradition" or "general agreement") we assume
: the root to be positive.
: If the negative root is desired, it's indicated
: it with a minus-sign: -sqrt(9). Otherwise, we
: must ASSUME it is the postive square root.
: (It's not just a good idea, it's the LAW!)
: About that problem: you derivative is correct.
: We set it equal to 0 and solve for x.
: -200x^(-2) - 5x^(-3/2) = 0
: -200/x^2 - 5/x^(3/2) = 0
: Multiply through by x^2:
: -200 - 5x^(1/2) = 0
: -5x^(1/2) = 200
: x^(1/2) = -40
: Square both sides: x = 1600
: We have a Critical Value!
: So what's the problem?
: Well, it doesn't check!
: Plug it into y', the DERIVATIVE.
: -200/(1600)^2 - 5/(1600)^(3/2)
: = -200/2560000 - 5/64000
: = -1/12800 - 1/12800 = - 1/6400
: It doesn't work! A critical value is supposed
: to make the derivative equal 0.
: When this happens, we'll find that it's always
: "that close" to working. (Or as Maxwell Smart
: used to say, "Missed it by THAT MUCH!")
: The second fraction has -5/(1600)^(3/2).
: The 3/2 power says sqrt(1600^3) or (sqrt1600)^3.
: Either way, the sqrt-sign is there already -
: it's not visible, but it's there. So, we're
: forced to say that the denominator is +64000.
: Note that, if we're allowed to CHOOSE the sign
: on the sqrt -- that is, that sqrt(1600) = -40 --
: it WOULD have worked. (That's what I meant by
: being "that close".) But we're not given that
: Here's a simpler example: sqrt(x) = -3
: Square both sides: x = 9
: We seem to have the answer. But do we?
: Let's check it.
: The equation would be: sqrt(9) = -3.
: True or false?
: If you're thinking, "Well, it depends...", then
: you can see the problem.
: IF we may consider both square roots (+3 and -3)
: and IF we are allowed to choose one, then it's
: true. That's a lot of IFs.
: My theory is that "they" decided that it's easier
: for us if we eliminate the choices and that we
: will always take a sqrt-sign to mean a positive
: Wow, this is long! Hope this is clear enough, KK.
: If you still have questions, keep on asking.
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