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#1 2015-09-10 12:49:19

mrpace
Member
Registered: 2012-08-16
Posts: 88

Prove this one??

Suppose that f is continuous on the closed interval [a, b], a < b, and that f(a) =
A, f(b) = B, A < B.
Suppose further that f is strictly increasing on [a, b] (i.e. for any x1, x2 ∈ I, x1 <
x2, impliesf(x1) < f(x2)).
Let C be a number between A and B. Show that there is exactly one value of x in
(a, b) such that f(x) = C.


I used the bisection process to try to explain why this must be the case but i'm not convinced that's what they're looking for....any ideas???

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#2 2015-09-13 03:15:51

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,049

Re: Prove this one??

Hi mrpace

You can prove that there is at least one number c in (a,b) such that f(c)=C by the Intermediate Value Theorem. To prove that there is only one such number, we will take any two numbers, c1 and c2, such that f(c1)=f(c2)=C. If c1 were less than c2, than f(c1) would be less than f(c2) which is not possible. Also, if c1 were greater than c2, then f(c1) would be greater than f(c2), which is impossible as well. Thus, c1 must be equal to c2, which proves that there is exactly one such number.


“Here lies the reader who will never open this book. He is forever dead.
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