Mean value Theorems: ECE GATE 2017: Calculus

August 25, 2016

GATE 2017 : Mean value theorems: calculus

Video Lecture on GATE 2017 : Mean value theorems:

calculus

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.

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

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