I'm still curious about the minimum value of the function:
f(x) = (x-2006)(x-2007)(x-2008 )(x-2009).
Wen Shih, what is the best way to handle this? Probably there is some quick and clever way to see the minimum value of the function but I can't find it?
The actual question is from playfulgirl here: http://www.flickr.com/photos/mabuthaodang/8515758304/in/photostream
Some things are obvious, like the min. value is at least 0. Also the function clearly can take negative values, e.g. choose x such that the first term is positive and the other three negative. E.g. x = 2006.5.
Also, the 2006, 2007, 2008, 2009 are not important. The question is equivalent to finding the minimum value of: f(x) = x(x-1)(x-2)(x-3). because the difference is just a horizontal translation of the graph along the x-axis.
We can do the question the long way, e.g. expand the function, differentiate, and set to 0, but I'm curious what is the short way.
Hi,
The minimum value of
f(x) = (x-2006)(x-2007)(x-2008)(x-2009)
is the same as the minimum value of f(x) = (x-1)(x-2)(x-3)(x-4).
I considered the graphing approach (due to its ease of visualising any pattern) and observed that the minimum value of the curve
y = (x-1)(x-2)(x-3)(x-4)
occurred at one point where the line
y = x-1
intersects the curve y = (x-1)(x-2)(x-3).
Solving, we obtain x = (5/2) +/- (sqrt(5)/2).
Take x = (5/2) - (sqrt(5)/2), since the curve
y = (x-1)(x-2)(x-3)(x-4)
is symmetrical about the line x = 5/2 and we arrive (after some algebra) at the minimum value of -1.
P.S. I think that the olympiad question setter typically develops interesting questions (like tha one above) that rely on some mathematical pattern or special mathematical behaviour.
Thanks.
Cheers,
Wen Shih
wenshih.wordpress.com
Wen Shih thanks. So it's an Olympiad question? I was trying to find a "short way" with the question but maybe there's none! If I use the online GC, it also does show two minimum points, both of them -1.
Hi,
Looks like one and tedious to solve if we were to use traditional methods :P
Cheers,
Wen Shih
wenshih.wordpress.com