7 State the Mean Value TheoremSolution The Mean Value Theore

Solution

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} \ eq 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.