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You are not logged in. #1 20051220 11:27:50
tricky problem Let f be the function defined by f(x) = x^3  x^2 4x + 4. The point (a,b) is on the graph of f, and the line tangent to the graph at (a,b) passes through the point (0, 8), which is not on the graph of f. Find a and b. A logarithm is just a misspelled algorithm. #2 20051220 12:00:37
Re: tricky problemSorry, I can't think of anything. I thought I had something, but it turned out it was wrong. Then I thought I had another something, but that was wrong too. Why did the vector cross the road? It wanted to be normal. #3 20051220 14:04:18
Re: tricky problemI guess its just a genuine badass problem. I'll give it another shot. Last edited by mikau (20051220 14:05:15) A logarithm is just a misspelled algorithm. #4 20051220 16:47:04
Re: tricky problem2x^3  x^2  12 = 0 "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #5 20051220 17:20:30
Re: tricky problemYep yep. But like I said, it is not imediatly apparent that it can be factored into that form. If it was, then we could solve the problem right away. It leaves no remainder and thus is a zero of the expression or function when the factor equals 0. Last edited by mikau (20051220 17:25:18) A logarithm is just a misspelled algorithm. #6 20051220 22:31:21
Re: tricky problemWhat is the question coz i might able 2 do it????????????? #7 20051220 22:37:44
Re: tricky problemUmmm ... the question is the bold bit at the top. "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #8 20051220 23:44:38
Re: tricky problemWell, no one said math should be immediately apparent "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #9 20051221 09:59:57
Re: tricky problemIs there a method to finding the (x2) factor of the cubic? Last edited by John E. Franklin (20051221 10:54:29) igloo myrtilles fourmis #11 20051221 13:19:37
Re: tricky problem"Or did Ricky divide by it because we had already known the root?" "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #12 20051221 21:21:18
Re: tricky problemSo it was basically trial and error, then. If it wasn't factorisable, you would have been there for quite a while. Why did the vector cross the road? It wanted to be normal. #14 20051222 02:37:33
Re: tricky problem
Sort of. Since the coefficients are all low numbers, you know the factors have to be low as well. So you can just try all the low integer factors 5 to 5 and see if they work. If those don't, you can be reasonably sure the factors aren't integers. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." 