The mean-value theorem is one of the most important theorems in calculus. However, despite its roles in the development of calculus itself, the mean-value theorem can be used in estimation-related problems. This article provides some examples of how it is applied in that area.

 

The Mean-Value Theorem

If f is a function that is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) then there exists a point ξ ∈ (a,b) such that:

f^{\: \prime} (\xi) = \frac{f(b) - f(a)}{b - a}

 

Example 1

Give an estimate of the difference between \arcsin{\frac{3}{5}} and \frac{\pi}{6}.

Answer

We know that \arcsin{\frac{1}{2}} = \frac{\pi}{6}. By the mean-value theorem, we can calculate the range of the difference between \arcsin{\frac{3}{5}}  and \arcsin{\frac{1}{2}}, that is the difference between \arcsin{\frac{3}{5}} and \frac{\pi}{6}. We therefore let f(x) = arcsin x by restricting  f to the “new” domain [\frac{1}{2},\frac{3}{5}]. Note that f is continuous on [\frac{1}{2},\frac{3}{5}] and f is differentiable on (\frac{1}{2},\frac{3}{5}). More specifically, f^{\: \prime} (x) = \frac{1}{\sqrt{1 - x^2}} for every x \in (\frac{1}{2}, \frac{3}{5}). Thus f satisfies the sufficient condition for the consequence in the mean-value theorem. Choose \xi \in (\frac{1}{2},\frac{3}{5}) such that f^{\: \prime} (\xi) = \frac{\arcsin {0.6} - \arcsin{0.5}}{0.6 - 0.5}. (The existence of such ξ is guaranteed by the mean-value theorem.) From this we get \arcsin{0.6} - \frac{\pi}{6} = \frac{1}{10 \sqrt{1 - {\xi}^2}}. Since \frac{1}{2} < \xi < \frac{3}{5}, it can be shown that \frac{1}{5 \sqrt{3}} < \frac{1}{10 \sqrt{1 - {\xi}^2}} < \frac{1}{8}. It follows that \frac{1}{5 \sqrt{3}} < \arcsin{0.6} - \frac{\pi}{6} < \frac{1}{8}. So, the answer is: \frac{\sqrt{3}}{15} < \arcsin{\frac{3}{5}} - \frac{\pi}{6} < \frac{1}{8}.

 

Example 2

Suppose that a > 0. Prove that \frac{a}{1+a} < \ln(1+a) < a.

Answer

Let a > 0. Consider the real-valued function f defined on the closed interval [0,a] where f(x) = ln (1+x) for every x ∈ [0,a]. It can be shown that f is continuous on [0,a] and f is differentiable on (0,a). Specifically, f^{\: \prime} (x) = \frac{1}{1 + x} for every x ∈ (0,a). Thus f satisfies the sufficient condition for the consequence in the mean-value theorem. Choose ξ ∈ (0,a) such that f^{\: \prime} (\xi) = \frac{\ln (1+a) - \ln (1+0)}{a - 0} . As a consequence, we have the following.

\frac{1}{1 + \xi} = \frac{\ln (1+a)}{a} …………………………………………………………………………………………………………………………………………….. (*)

Since 0 < ξ < a, then 1 < 1 + ξ < 1 + a and this gives:

\frac{1}{1+a} < \frac{1}{1 + \xi} < 1 …………………………………………………………………………………………………………………………………………………………………………………………………………. (**)

Substituting (*) into (**), we get:

\frac{1}{1+a} < \frac{\ln (1+a)}{a} < 1

\frac{a}{1+a} < \ln (1+a) < a

 

Example 3

Prove that if n > N2 then \sqrt{n + 1} - \sqrt{n} < \frac{1}{2N} where n and N are natural numbers)

Answer

Let n > N2. Let f be the function on the closed interval [n,n+1] to \mathbb{R} given by f(x) = \sqrt{x} for every x ∈ [n,n+1]. Note that f is continuous on [n,n+1] and f is differentiable on (n,n+1). Specifically, f^{\: \prime} (x) = \frac{1}{2 \sqrt{x}}. Thus f satisfies the sufficient condition for the consequence in the mean-value theorem. Choose ξ ∈ (n,n+1) such that f^ {\: \prime} (\xi) = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n}. Consequently, \sqrt{n+1} - \sqrt{n} = \frac{1}{2 \sqrt{\xi}}. Since n < ξ < n+1, it holds that \frac{1}{2 \sqrt{n+1}} < \frac{1}{2 \sqrt{\xi}} < \frac{1}{2 \sqrt{n}}. From the assumption that n > N2 it can be shown that \frac{1}{2 \sqrt{n}} < \frac{1}{2N}. As a consequence, \sqrt{n+1} - \sqrt{n} < \frac{1}{2N}.

 

Example 4

Suppose that we approximate \sqrt{101} by 10. By applying the mean-value theorem, give an upper bound of the difference between the exact value of \sqrt{101} and 10.

Answer

Note that 101 and 100 differ by 1. Thus we can apply the proposition in Example 3. Select n = 100 and N = 9 Note that 100 > 92.  By the proposition, we can be sure that \sqrt{101} - \sqrt{100} = \sqrt{101} - 10 < \frac{1}{2 \cdot 9} < .056. Thus, the difference between the exact value of \sqrt{101} and 10 is less than .056.

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