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 January 4th, 2015, 08:59 PM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 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.
 January 4th, 2015, 09:58 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,805 Thanks: 1045 Math Focus: Elementary mathematics and beyond 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.
 January 4th, 2015, 10:16 PM #3 Newbie   Joined: Jul 2012 From: Houston, Texas Posts: 18 Thanks: 2 Math Focus: Geometry , Modern Algebra 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.
 January 5th, 2015, 05:11 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 Math Focus: Mainly analysis and algebra Strictly speaking, this is Bolzano's theorem which is a special case of the intermediate value theorem.

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# tanx=2x(0,1.4)

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