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Old 2005 March 28th, 01:05 PM   #21 (permalink)
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Re: Intersection

The answer is 3.

1. For x < 0,
- x^12 goes from +infinity to 0
- 2^x goes from 0 to 1
so the 2 curves must intersect at one point, just use the Newton theorem that helps to find zero point.
2. For 0<x<2
- for x = 0, 0^12 = 0 and 2^0=1, so 2^x is above x^12 for x = 0
- for x = 2, 2^12 = 4096 and 2^2=4, so x^12 is above 2^x for x = 2
so the 2 curves must intercept at one point between 0 and 2
3. For x>2, x-> infinity
lim 2^x > lim x^12 when x-> infinity (a simple rule says that exponential goes faster to the infinity than any x^n)
so the 2 curves must intercept at one point between 2 and the infinity.

Definitely the answer is D.Three(3).

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Old 2007 May 11th, 11:38 AM   #22 (permalink)
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I thought of it the same way as dragoninfinity/matroid (using concept of function growth order). Anyways, I am interested in the algebraic solution, so can someone please provide one (in the spirit of mathematical exactness )
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Old 2008 December 18th, 11:45 PM   #23 (permalink)
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Think roughly about the shape of the graphs. There is one intersection with
x < 0. Then when x= 0, x^12 is lower than 2^x. When x = 2, x^12 is
higher. But we know that for very large x ("in the limit") x^12 ends up lower.
So (looking at the general shape of the graphs) there must be two
intersections with x > 0. So there are three intersections altogether.
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