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i've attempted two questions i had, and if someone could post any feedback, it'd be great.
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
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2. x^5 - 3x^4 - 2x^3 - x + 1 = 0 has atleast one solution between 0 and 1.
let f(x) = x^5 - 3x^4 - 2x^3 - x + 1, where x E [0,1]
f(0) = 1
f(10) = -4
so..
f(1) < f(x) < f(0)
-4 < f(x) < 1
as f(x) has a min and max of -4 and 1 respectively for x E [0,1], the function attains every value between [-4,1] atleast once as f(x) is a continuous function over the interval x E [0,1]
therefor f(x) = x^5 - 3x^4 - 2x^3 - x + 1 has atleast one solution over x E [0,1]
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any thoughts on where i'm not clear in the proof? the odd part was in the first one, there was no interval restriction. did i do right in restricting it to a given interval? otherwise i cant see any method of showing it.
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