Limits+-+Intermediate+Value+Theorem


 * Intermediate Value Theorem

Property: //The Intermediate Value Theorem Textbook definition://** If //f(x)// is a continuous function on [//a,b//], for every //d// between //f(a)// and //f(b)//, there exists a //c// between //a// and //b// so that //f(c)// = //d//.

Let’s say //f// is a continuous function on a closed interval with the domain [//a,b//] as shown above. The intermediate value theorem shows you can take any //y// value between //f(a)// and //f(b)//, such as //d//, and there will be a value, //x// = //c//, in between //a// and //b// where //f(c)// = //d//. Essentially, this theorem is saying that, with these conditions, //d// will cross the function at least once. And, as you can see, there is a value //f(c)// = //d//. In this case there is only one //c// value, however there can be more if the function oscillates.
 * In Other Words**:

Use the IVT to prove that the polynomial, **x** **5** **+4x** **2** **-2x+3** has one real root. f(x)=x 5 +4x 2 -2x+3 has a root if there is a value of x that makes this statement, x 5 +4x 2 -2x+3=0, true. Let a=-10, f(-10)=(-10) 5 +4(-10) 2 -2(-10)+3=-100,000+400+20+3=-99,577<0 Let b=2, f(2)=(2) 5 +4(2) 2 -2(2)+3=32+16-4+3=47>0 Could we have used other values of a and b? If you are not given the interval there can be many possible appropriate intervals you can choose from. f is continuous on the interval [-10,2] because polynomial functions on their domains and the domain of a polynomial is the set of all real numbers. f(x)=x 5 +4x 2 -2x+3 is continuous on interval [-10,2] f(-10)=-99,577<0 and f(2)=47>0: f(-10)<0<f(2) We want to use the Intermediate Value Theorem to show that f(x)=0 somewhere between x=-10 and x=2 and N=0 is between f(-10) and f(2) then there must be a point c in (-10,2) such that f(c)=N=0
 * Example:**
 * Step 1: Create an appropriate function: f(x)=x 5 +4x 2 -2x+3**
 * Step 2: Find appropriate endpoints for an interval [a,b] so that when you evaluate f(a) and f(b) one value is greater than 0 and the other is less than 0.**
 * Step 3: Show that f is continuous on the interval.**
 * Step 4: Put the pieces together into an understandable explanation using the Intermediate Value Theorem.**

Bob measures his height one day (day 0) and he is 150 cm tall. Bob then measures his height 100 days later and finds he is now 160 cm tall. If we were to graph his height as a function of every day of this year, the function would naturally be continuous.
 * Real World Example:**

The closed interval would have these boundaries: //f(a) = f(0 days) = 150 cm// //f(b) = f(100 days) = 160 cm// Using such a graph, you could be asked to find the date when Bob was exactly 156 cm tall. This scenario satisfies the hypotheses of the intermediate value theorem; Bob’s growth is continuous over the closed interval of day 0 to day 100, and there is a value //d =// 156 cm between //f(0) = 150// and //f(100) = 160//. According to the intermediate value theorem any height between 150 cm and 160 cm can be chosen and there will be a certain day in this closed interval where Bob was this height.


 * Real World Application 2: Stabilizing a wobbly table**

This works on all 4 legged tables where the legs are equal in length and the instability is being caused by unevenness in the floor.

A table with 4 even legs will always rest on 3 legs. Picture a square table with corners/legs labeled A, B, C, and D in clockwise order starting with A in the NW corner. In the original position legs B, C, and D are touching the ground and leg A is elevated above the ground, thus, a wobbly table. Let's say that leg A is 1 inch above the ground. To make leg A come in contact with the ground you could press really hard on the A-B edge and force leg A to touch the ground while also forcing leg B into the floor (at this point leg B would be 1 inch below the floor); as this is not usually a feasible option, you could, instead, rotate the table 90 degrees in the clockwise direction while keeping legs B, C, and D flat on the ground. This full rotation would, in turn, force leg A into the floor (at this point leg A would be 1 inch below the floor) - much like what happened to leg B when the entire side was pushed. As leg A started above the floor and after the rotation ended below the floor, by Intermediate Value Theorem, at some point in the rotation leg A had to be even with the floor. At this specific point all legs will touch the ground and the wobble has been fixed.

f(a) = f(before rotation) = +1 inch f(b) = f(after 90 degree rotation clockwise) = -1 inch

The height of leg A is a continuous function as it has a definite solution for each point in the rotation. So, there must be some point c such that f(c) = 0 on the closed interval of [a,b].

Got it? Good.

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 * If you are still confused, here is a helpful link that goes over the same material:

-[| http://www.calculus-help.com/funstuff/tutorials/limits/limit06.html] (Images 1, 2, and 3) -[| http://www.coolmath.com/graphit/] (Image 4) - Foerster, Paul A. //Calculus: Concepts and Applications//. 2nd ed. Key Curriculum, 2005. Print. http://74.125.95.132/searchq=cache:CFNg2AYfMNYJ:asmsa.org/math/marizza/Calculus/IVTexamples.doc+intermediate+value+theorem+to+prove+that+the+polynomial,+x+5%2B4x2-2x%2B3+has+one+real+root.&cd=1&hl=en&ct=clnk&gl=us
 * Sources**

- [|http://www.maa.org/devlin/devlin_02_07.html]