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January 4th, 2015, 08:59 PM   #1
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Intermediate Value Theorem

Use the Immediate Value Theorem to show that there is a root of the given equation in the specified interval.

tan(x)=2x (0,1.4)

How to solve this question?

Last edited by greg1313; January 4th, 2015 at 10:02 PM.
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January 4th, 2015, 09:58 PM   #2
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It's intermediate value theorem, not "Immediate Value Theorem".

Since one can find values where tan(x) - 2x is negative and where tan(x) - 2x is positive in the
given interval, at some point in the given interval tan(x) - 2x = 0 (by the intermediate value theorem).

I don't know how much detail is required to provide an acceptable answer your question.
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January 4th, 2015, 10:16 PM   #3
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I will give a try.
Take the function $f(x) = \tan(x) - 2$
We know that $f(x)$ is continuous on the interval $(0,1.4)$
Because this is an open interval, we can't choose 0 and 1.4. Pick two value x= $0.1$ and $x =1.39$ to test the sign of $f(x)$.
At $x = 0.1, f(0.1) < 0$
At $x = 1.39, f(1.39) > 0$
The Intermediate Value Theorem states that if $f(a).f(b) <0$ then there exist an $x$ between$ (a,b)$ such that $f(x) = 0.$
The problem can also be solve by using Intermediate Value Theorem for Derivative.

Last edited by skipjack; January 5th, 2015 at 06:10 AM.
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January 5th, 2015, 05:11 AM   #4
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Strictly speaking, this is Bolzano's theorem which is a special case of the intermediate value theorem.
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