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# minimum value of function

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How can I find the minimum value of f(x) = x^2 -2mx? I know that when x = m, its at its lowest (differentation).

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Am I missing something?

You say you know the lowest value is at x=m, so the lowest value is f(m), although I'm sure you must mean something else.

Also differentiation isn't the best way, why not just complete the square?

[edited by - higherspeed on April 11, 2004 9:51:44 AM]

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It''s not even necessary to complete the square:

x^2 - 2mx = 0
x(x - 2m) = 0
x = 0 V x = 2m

Minimum value is at x_min = (0 + 2m)/2 = m

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AP: that solution doesn''t show it''s the lowest point, why not just quote the value m? That working is pointless and does nothing to explain why the answer is correct.

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The min or max value of a function f(x)=ax^2+bx+c is at x0=(x1+x2)/2, where f(x1)=f(x2)=0 (which is easily proven through differentation, for example). f(x0) is the minimum value if a>0 and maximum if a<0 (again, easy to prove). And after you''ve proven it once you can start using the result without proving it again every time.

BTW, how would you actually do this by just completing the square and not using this method..? I thought you meant something like this:
x^2-2mx=0
(x-m)^2=m^2 (this is called completing the square, correct?)
x-m=+-m
x=m+-m

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The best way to provide that this is a local minimum is to take two derivatives.

f(x) = x^2-2mx
f''(x) = 2x-2m
f''''(x) = 2

From f''(x) you can show that it is zero at x=m which corresponds to either a local minima or maxima. f''''(x) gives you the concavity of the function. Since f''''(x) is a constant positive value f(x) is said to be concave-up (like a cup). Concave-up indicates a minumum at x=m rather than a maxima. Additionally since f''''(x) is constant you can prove that x=m is the absolute minumium for f(x).

-Dan

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Ok, differentiation is just taking a relatively advanced tool to solve an extremely simple problem.

(x - m)^2 >= 0 Simple to prove for any totally ordered field.

so (x - m)^2 - m^2 >= -m^2
which is the same as x^2 - 2xm >= -m^2

now put in x = m to show that -m^2 does belong to the range of f, so the least value of f is -m^2.

This solution is very simple and very rigorous. It does not require any specific results for quadratic functions.

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quote:
Original post by higherspeed
Ok, differentiation is just taking a relatively advanced tool to solve an extremely simple problem.

No, it's taking a tool that works well for this type of problem and applying it, however "advanced" it may be or not. The hardest part of this problem would probably be differentiating x2 and 2mx (with respect to x) in your head, which literally doesn't take any time at all.

Not to say your method isn't as valid as differentiation, it's just that you seem to be driving the point that differentiation isn't the best way of solving this problem, whereas I see finding local extrema of functions one of the fundamental applications of differentiation - at least one of the first things I learned to do with it - regardless of the inherent simplicity or form of any such function. Even though the example takes on a common form for which there are other methods of finding extrema, differentiation is the general solution. And even given these other methods, it's questionable whether or not they are always "better" than differentiation. And let's not forget that smart feeling we all get when we use Calculus to solve problems, especially the ones with which we were stuck using older methods back in our early algeabra days

[edited by - Zipster on April 12, 2004 3:34:53 AM]

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quote:
Original post by higherspeed
(x - m)^2 >= 0 Simple to prove for any totally ordered field.

Why do you care about finding a formula for minimum solutions to a quadratic when derivatives solve the general problem?

[edited by - sjelkjd on April 12, 2004 5:21:27 AM]

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This isn''t the general problem, it''s a particular problem, also differentiation doesn''t even come close to solving the general problem. It can solve it for differentiable, well-behaved functions only, for a general function, which doesn''t have to be continuous even, it can be next to useless.

Differentiation is fine when it can be applied, as it can be in this case. I was just giving a far nicer solution. The thing that I have most issue with was the AP''s solution, which was right but had no justification and relied on results that most people would not bother knowing, even if they are obvious, it was just taking a very long way round a very simple problem.

Solutions like that simple serve to cloud people''s understanding in my opinion, as does constantly using differentiation, without ever thinking about the problem. It leaves people stuck when a problem arises when differentiation is useless.

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