Re: More Challenges Message #6 Posted by David Hayden on 2 Feb 2009, 11:09 p.m., in response to message #5 by Chuck
Okay, my calculus is really rusty, but here's what I came up with. There's a good chance that I've gotten something horribly wrong, in which case, corrections will be cheerfully accepted.
To keep my head from hurting, I recast the problem:
Let u(x) = 1/x
Let g(u) = floor(u)*u
Then f(x) = floor(1/x)/x is the same as g(u(x))
If N is a negative integer then For u = ( N , N+1 ], floor(u) is the constant N+1 and g(u) = (N+1)*u.
Inside one of these intervals, the derivative of f(x) is:
f'(x) = g'(u) * u'(x) = (N+1) * -1/x^2
= -(N+1) / x^2
= -(N+1) * u^2
Now we can find where f'(x)=3:
3 = -(N+1) * u^2
u = sqrt(-3/(N+1)) or -sqrt(-3/(N+1))
Since we already said that N is a negative integer and u is in the interval (N, N+1], that means u is a negative number also, so we can get rid of the positive solution and we're left with:
u = -sqrt(-3/(N+1))
Now lets try some solutions. N=-1 won't work because -3/(N+1) causes division by zero.
How about N=-2?
u = -sqrt(-3/(-1))
= -sqrt(3)
= -1.732050...
Checking our assumption:
N <? u <=? N+1
-2 < -1.732050... <= -1
That works.
Last but not least, we solve for x:
u = 1/x
-sqrt(3) = 1/x
x = -1/sqrt(3)
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