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Old 04-12-2004, 01:36 PM   #8 (permalink)
iccky
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Location: Princeton, NJ
Quote:
Originally posted by frozenstellar
1. If f(x) = x^3 - 5x^2 + 7x - 9, prove there is a real number c such that f(c) = 100.

let g(x) = x^3 - 5x^2 + 7x - 109 where g(c) = 0.

when x = 0, g(0) = -109
when x = 10, g(10) = 461

let x E [0,10] (did this as there was no interval restiction stated in the question)

g(0) < g(x) < g(10)
-109 < g(x) < 461

as g(x) is continuous over x E [0,10], and the min and max over this domain are -109 and 461 respectively, by the IVT the function must attain every value in [-109, 461] atleast once.

so g(c) = 0
and therefor f(c) = 100, where c is a real number
As Yak said, not quite right on number 1. The easiest way to do this, if I remember correctly, is to choose an interval, find teh min and max on that interval and then state that G(x) is between that min and max (this is true, whereas saying g(x) is between 0 and 10 because x is between 0 and 10 is not necessarily true).

Its been awhile though.
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