Posted by Subhotosh Khan on August 28, 2002 at 08:31:58:
In Reply to: Re: minimising and maximising posted by KK on August 28, 2002 at 01:52:43:
: : : :But how does it prove that the equation
: : : : y=(2000+20x+10x^1/2)/x cannot be minimised (i.e. the function does not have a convex). I thought we have to take the second derivative because for a convex to occur f"(x)>0.
: : : Please listen carefully, KK.
: : : The function had NO critical values.
: : : (We set the first derivative equal to 0
: : : and could not solve for x.)
: : : Therefore, there are NO maximums or minimums.
: : : Why do we take the second derivative?
: : : To test the Critical Values for max or min.
: : : But there are NO critical values to test.
: : : [Your flashlight doesn't work. I tell you that
: : : there are no batteries in it. And you insist on
: : : trying a battery-tester?]
: Hi, sorry I know that I'm quite slow at maths, but I have tried drawing the graph on Microsoft excel. I got a graph with no miniums :) .. but the graph shows that at zero, it is undefined. I.e. When x=0.009 the corresponding y was a huge number.
: It falls and falls but never increase again. Is that part counted as maximising ? IF it is a maximising part, then how come there's no critical value for it?
: Thank you so much for your help. I really appreciate it because uI have gained so much more understanding by asking those questions.. Thank you for your time.
The function has an ABSOLUTE minimum at x =0. However, at that point y' = (undefined)....(NOT y'=0.. no battery in flashlight).
So you CANNOT use the y" < > 0 for maxima-minima test (don't need battery tester).
You have the right approach of INVESTIGATING the curve at the limits through spreadsheets - I always do it that way.
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