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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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quote:
Original post by higherspeed
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.

It''s the best general solution I can think of, even if it doesn''t work in a lot of cases. At which point specific knowledge about the function is necessary.

quote:
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.

I don''t really think your solution was "far nicer." Knowing that completing the square gives you something where you know (x - m)^2 >= 0 (for which the minimum is the obvious from there) is on par with knowing there''s an extreme when the derivative is zero. I guess it''s simply more natural for me to differentiate since I''ve been in Calculus courses for the past three years (and counting).

quote:
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.

I guess you''re referring to the AP''s solution, which I didn''t really pay attention to since it wasn''t very fleshed out - so I can''t comment on that. But I really don''t think this is an issue of people not being properly equipped to handle problems when differentiation doesn''t work. I for one had to take several years of algebra before taking any Calculus courses, and conic sections in particular were pounded into our heads, along with all sorts of other types of equations (logarithmic, transcendental, etc.). I learned them, and understood them. Then I moved on to the world of Calculus, which made a lot of the earlier problems a lot easier to work with. And thus I get away with using it when I can. But if I was stuck with an ill-behave, discontinuous function, I''d fall back on the basics.

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

What I cant see is that if that is the minimum value then shouldnt there only be one value (bottom of quadratic curve)?
Using differentation-
x^2 - 2mx

dy/dx = 2x - 2m
Lowest value is when grad = 0
2x - 2m = 0
x = m

f(m) = -m^2 (because variable x = m at this point)

Hmmmm, Is this the minimum (think ive done overkill :S)

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I suppose which solution is nicer depends on whether you look at it from a pure, or applied perspective. I''m used to mainly pure maths, only a quarter of my uni course so far is to do with the methods of calculus and my supervisor for half of that specialises in groups.

As a result when asked questions I try to give an extremely rigorous answer and I know which solution is easiest to justify from that point of view.

Genjix: the first 3 lines you''ve written in the last post serve to show that two positions where the function is zero are at 0 and 2m, then using the fact that a parabola is symmetric, the minimum must be in the middle of these points, ie at m. However this method isn''t very nice, so stick with your solution using differentiation. Or if you want to understand an algebraic solution look at the solution using completing the square.

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ah huh, enlightened me makes sense now.
x = m => 2x - 2m = 2x - 2x
= 0
so the gradient is 0 when m = x (flat part of quadratic curve)