3 min readβ’february 15, 2024

The** first condition** states that *f *must be **continuous **on the closed interval [a, b]. From our first unit, we know that that simply means that there are **no holes, asymptotes, or jump discontinuities in the graph** between points *a* and *b*. Because they are closed brackets, the graph must be continuous at the points *a* and *b.Β *

The **second condition** states that Β *f *must be **differentiable **on (a, b). Note that this time, itβs an open interval. An equation is differentiable at *a* if it is **continuous at ****a** and if **lim x->a [f(x) - f(a)]/(x - a) exists**. Most continuous equations are differentiable unless they have a corner (as in an absolute value function)Β

If we meet these two conditions, then we can conclude that there exists a point *c *on (a, b) such that f'(c) = [f(b)-f(a)]/b-a. In other words, **there is a point where the slope of the tangent line is equivalent to the slope of the secant line between ****a**** and ****b.****Β ****π**

From the Mean Value Theorem, we can derive **Rolleβs Theorem**, which simply states that if f(a) and f(b) are equal to each other, then there will be some point on the graph where the slope of the tangent line is equal to 0. β

Example 1:
Consider the function f(x) = x^3 on the interval [0, 1]. To find the mean value, we use the formula f'(c) = (f(b) - f(a))/(b-a), where c is a value in the interval [a, b]. Plugging in a = 0, b = 1, and f(0) = 0 and f(1) = 1, we get f'(c) = 1. So, the mean value of f(x) = x^3 on the interval [0, 1] is 1.

Example 2:
Consider the function g(x) = x^2 on the interval [-1, 1]. To find the mean value, we use the formula g'(c) = (g(b) - g(a))/(b-a), where c is a value in the interval [a, b]. Plugging in a = -1, b = 1, and g(-1) = 1 and g(1) = 1, we get g'(c) = 0. So, the mean value of g(x) = x^2 on the interval [-1, 1] is 0.

Example 3:
Consider the function h(x) = cos x on the interval [0, Ο/2]. To find the mean value, we use the formula h'(c) = (h(b) - h(a))/(b-a), where c is a value in the interval [a, b]. Plugging in a = 0, b = Ο/2, and h(0) = 1 and h(Ο/2) = 0, we get h'(c) = -2/Ο. So, the mean value of h(x) = cos x on the interval [0, Ο/2] is -2/Ο.

Example 4:
Consider the function k(x) = sin x on the interval [0, Ο/2]. To find the mean value, we use the formula k'(c) = (k(b) - k(a))/(b-a), where c is a value in the interval [a, b]. Plugging in a = 0, b = Ο/2, and k(0) = 0 and k(Ο/2) = 1, we get k'(c) = 2/Ο. So, the mean value of k(x) = sin x on the interval [0, Ο/2] is 2/Ο.

The Mean Value Theorem states that for any continuous function on a closed interval, there exists a value c in the interval such that the value of the derivative of the function at c is equal to the average rate of change of the function over the interval. By using this theorem, we can find the mean value of a function on a given interval, which can provide useful information about the behavior of the function.

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