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Mean Value Theorem
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By Rick Stoll          

 Calc I

 

[Under Construction]

 

 

 

The Mean Value Theorem is one of the most important theoretical tools in Calculus. It states that 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

 

\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a} \cdot\end{displaymath}

 

The special case, when f(a) = f(b) is known as Rolle's Theorem. In this case, we have f '(c) =0. In other words, there exists a point in the interval (a,b) which has a horizontal tangent. In fact, the Mean Value Theorem can be stated also in terms of slopes. Indeed, the number

 

\begin{displaymath}\frac{f(b) - f(a)}{b-a} \end{displaymath}

 

is the slope of the line passing through (a,f(a)) and (b,f(b)). So the conclusion of the Mean Value Theorem states that there exists a point $c\in(a,b)$ such that the tangent line is parallel to the line passing through (a,f(a)) and (b,f(b)). (see Picture)

 

 

Example. Let $f(x) = \displaystyle \frac{1}{x}$, a = -1and b=1. We have

 

\begin{displaymath}\frac{f(b) - f(a)}{b-a} = \frac{2}{2} = 1.\end{displaymath}

 

On the other hand, for any $c \in(-1,1)$, not equal to 0, we have

 

\begin{displaymath}f'(c) = - \frac{1}{c^2} \neq 1.\end{displaymath}

 

So the equation

 

\begin{displaymath}f'(c) = \frac{f(b) - f(a)}{b-a}\end{displaymath}

 

does not have a solution in c. This does not contradict the Mean Value Theorem, since f(x) is not even continuous on [-1,1].

Remark. It is clear that the derivative of a constant function is 0. But you may wonder whether a function with derivative zero is constant. The answer is yes. Indeed, let f(x) be a differentiable function on an interval I, with f '(x) =0, for every $x \in I$. Then for any a and b in I, the Mean Value Theorem implies

 

\begin{displaymath}\frac{f(b) - f(a)}{b-a} = f'(c)\end{displaymath}

 

for some c between a and b. So our assumption implies

 

\begin{displaymath}f(b) - f(a) = 0 \cdot(b-a) = 0.\end{displaymath}

 

Thus f(b) = f(a) for any aand b in I, which means that f(x) is constant.

Exercise 1. Show that the equation

 

2x3 + 3x2 + 6x + 1 = 0

 

has exactly one real root.

 

 

 

Exercise 2. Show that

 

\begin{displaymath}\vert\cos(2a) - \cos(2b)\vert \leq 2\vert a-b\vert\end{displaymath}

 

for all real numbers a and b. Try to find a more general statement.

 

 

 

 

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Last modified: April 30, 2001