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.

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)

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.

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)