Uniqueness+Theorem,+Mean+Value+Theorem,+Rolle's+Theorem

Uniqueness Theorem, Mean Value Theorem, Rolle's Theorem Definition: If: 1. //f'(x) = g'(x)// for all values of //x// in the domain, and 2. //f(a) = g(a)// for one value, //x = a,// in the domain, then //f(x) = g(x)// for all values of //x// in the domain.
 * //I. Uniqueness Theorem//**


 * // II. Mean Value Theorem - if // f(x) // is defined and continuous on the interval [a,b] // and differentiable on // (a,b), then there is at least one number c in the interval (a,b) (that is a<c<b) such that: f'(c)= [f(b)-f(a)]/(b-a) //**

As you can see in the graph above, the secant line between points //a// and //b// and the tangent line at point //c// are parallel or they have equal slopes.

__Sample Problem__ You may be asked to find a value of //c// that satisfies this theorem, such as with **f(x) = -2x 3 + 6x - 2** on the interval [-2, 2]. The graph is below. First find the slope of the secant line that crosses through the endpoints of the interval. f(-2) = -2(-2) 3 + 6(-2) - 2 = 2 f(2) = -2(2) 3 + 6(2) - 2 = - 6 To find the slope plug these values into **// [f(b)-f(a)]/(b-a). //** [f(b) - f(a)] / (b - a) = [ -6 - 2 ] / (2 - -2) = -2 Since the slope of the secant line is -2, the slope of the tangent line must also be -2. To find the slope of the tangent line set the derivative of the function equal to this value. f '(x) = -6x 2 + 6 -6c 2 + 6 = -2 c = 2 sqrt(1/3) and c = - 2 sqrt(1/3) The mean value theorem is satisfied at two values of c in this example, however there should always be one value of c where **// f'(c)= //****// [f(b)-f(a)]/(b-a). //**

An example of a function that satisfies this theorem is below.
 * // III. Rolle's Theorem - Rolle's Theorem is just like the Mean Value Theorem except for it is the special case, when f(a)=f(b). We have f'(c)=0. In other words, there exists a point in the interval (a,b) which has a horizontal tangent. //**

__Sample Problem__ We can find the point //c// that satisfies Rolle's Theorem on the simple graph of //y// = //x//2 − 2 on the interval [- √2, √2]. The endpoints of the interval are the zeros of the function as can be seen below. To find //c// set the derivative of y equal to zero. //y' = 2x = 0 x = 0// = //c// As you can see the value of //c// that does satisfy Rolle's Theorem is x = 0.

Please go to survey :) []

Bibliography:
 * []
 * http://www.maa.org/joma/Volume8/Kalman/Rolle.gif
 * http://www.analyzemath.com/calculus/MeanValueTheorem/meanvaluetheorem.gif
 * []

Topic: Rolle’s, MVT and Uniqueness Theorem Group Members: Anna and Katherine

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Hey, what’s that whole paragraph underneath the Uniqueness Thm. Looks like you just copy pasted without looking at what it was saying. It’s talking about partial derivatives with respect to y. So the Uniqueness Thm needs a new explanation and possibly a graph demonstrating it. Your notation needs improvement. Don’t use ^ and sqrt. Nice organization and graphs in the last two. Could be a bit more creative.

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