4.4: The Mean Value Theorem

The difference between instantaneous and average rates of change has been discussed in earlier sections. Recall that there is no difference between the two for linear functions. For nonlinear functions the average rate of change over an interval \(\ival{a}{b}\) of positive length (i.e. \(b-a>0\)) will not be the same as the instantaneous rate of change at every point in the interval. However, the following theorem guarantees that they will be the same at some point in the interval:

Figure [fig:mvt] below shows the geometric interpretation of the theorem:

The idea is that there is at least one point on the curve \(y=f(x)\) where the tangent line will be parallel to the secant line joining the points \((a,f(a))\) and \((b,f(b))\). For each \(c\) in \((a,b)\) the tangent line has slope \(f'(c)\), while the secant line has slope \(\frac{f(b) - f(a)}{b - a}\). The Mean Value Theorem says that these two slopes will be equal somewhere in \((a,b)\).

To prove the Mean Value Theorem (sometimes called Lagrange’s Theorem), the following intermediate result is needed, and is important in its own right:

Figure [fig:rolle] on the right shows the geometric interpretation of the theorem. To prove the theorem, assume that \(f\) is not the constant function \(f(x) = 0\) for all \(x\) in \(\ival{a}{b}\) (if it were then Rolle’s Theorem would hold trivially). Then there must be at least one \(x_0\) in \((a,b)\) such that either \(f(x_0) > 0\) or \(f(x_0) < 0\). If \(f(x_0) > 0\) then by the Extreme Value Theorem \(f\) attains a global maximum at some \(x=c\) in the open interval \((a,b)\), since \(f\) is zero at the endpoints \(x=a\) and \(x=b\) of the closed interval \(\ival{a}{b}\). Then \(f'(c)=0\) since \(f\) has a maximum at \(x=c\). Likewise if \(f(x_0) < 0\) then \(f\) attains a global minimum at some \(x=c\) in \((a,b)\), and thus again \(f'(c)=0\;.\quad\checkmark\)

The Mean Value Theorem can now be proved by applying Rolle’s Theorem to the function

\[F(x) ~=~ f(x) ~-~ f(a) ~-~ \frac{f(b) - f(a)}{b - a}\,(x - a) \nonumber \]

where \(f\) satisfies the conditions of the Mean Value Theorem. Basically, \(F\) “tilts” the graph of \(f\) from Figure [fig:mvt] to look like the graph in Figure [fig:rolle]. It is trivial to check that \(F(a) = F(b) = 0\), and \(F\) is continuous on \(\ival{a}{b}\) and differentiable on \((a,b)\) since \(f\) is. Thus, by Rolle’s Theorem, \(F'(c) = 0\) for some \(c\) in \((a,b)\). However,

\[F'(x) ~=~ f'(x) ~-~ \frac{f(b) - f(a)}{b - a} \nonumber \]

and so

\[F'(c) ~=~ f'(c) ~-~ \frac{f(b) - f(a)}{b - a} ~=~ 0 \quad\Rightarrow\quad f'(c) ~=~ \frac{f(b) - f(a)}{b - a} \quad\checkmark \nonumber \]

Note that both the Mean Value Theorem and Rolle’s Theorem are purely existence theorems—they tell you only that a certain number exists. The task of finding the numbers is left to you. For the Mean Value Theorem that task involves solving the equation \(f'(x) = \frac{f(b) - f(a)}{b - a}\) (or \(f'(x)=0\) for Rolle’s Theorem). The numerical root-finding methods from Section 4.3 could come in handy, since obtaining closed-form solutions might be impossible. For that reason, the Mean Value Theorem is more useful for theoretical purposes. One such application is the following important result:

Note that \(I\) can be any interval, even the entire real line \((-\infty,\infty)\). It is already known that \(f=\text{constant} ~\Rightarrow~ f'=0\); the above result says that the converse is true. The proof is by contradiction: assume that \(f\) is not a constant function and show this contradicts the Mean Value Theorem. If \(f\) is not constant then there exist numbers \(a < b\) in \(I\) such that \(f(a) \ne f(b)\). However, by the Mean Value Theorem there must exist a number \(c\) in the interval \((a,b)\) such that

\[f'(c) ~=~ \frac{f(b) - f(a)}{b - a} ~. \nonumber \]

Since the derivative of \(f\) is 0 everywhere in \(I\), then \(f'(c) = 0\) and so

\[\frac{f(b) - f(a)}{b - a} ~=~ 0 \quad\Rightarrow\quad f(b) ~-~ f(a) ~=~ 0 \quad\Rightarrow\quad f(a) ~=~ f(b) ~, \nonumber \]

a contradiction of \(f(a) \ne f(b)\). Thus, \(f\) must be a constant function.

Another theoretical result can be proved with the Mean Value Theorem:

To prove part (a), assume that \(f'(x) > 0\) for all \(x\) in \(I\), and choose arbitrary numbers \(a\) and \(b\) in \(I\) with \(a < b\). To prove that \(f\) is increasing on \(I\) it suffices to show that \(f(a) < f(b)\). By the Mean Value Theorem there is a number \(c\) in \((a,b)\) (and hence in \(I\)) such that

\[\begin{aligned} \frac{f(b) - f(a)}{b - a} ~&=~ f'(c) \quad\text{, and so}\

\[6pt] f(b) - f(a) ~&=~ (b-a)\,f'(c) ~>~ 0\end{aligned} \nonumber \]

since \(b-a>0\) and \(f'(c)>0\). Thus, \(f(b) > f(a)\), and so \(f\) is increasing on \(I.\quad\checkmark\)

The proof of part (b) is similar and is left as an exercise. You might wonder why such a proof is necessary. After all, an intuitive explanation was provided in Section 1.2 for why positive or negative derivatives imply that a function is increasing or decreasing, respectively. That knowledge has been assumed and used in the subsequent sections. Intuitive so-called “hand-waving” explanations, in fact, often yield more insight than a “formal” proof, such as the one above. However, it is good to know that such intuition has a solid basis and can be proved, if needed. The Mean Value Theorem can help in proving inequalities, often used in the sciences for establishing upper or lower bounds on a quantity (e.g. worst-case scenario).

Example \(\PageIndex{1}\): ineq1

Add text here.

Solution

Show that \(\sin\,x \le x\) for all \(x \ge 0\).

Solution: The inequality holds trivially for \(x=0\), since \(\sin\,0 = 0 \le 0\). So assume that \(x > 0\). Then by the Mean Value Theorem there is a number \(c\) in \((0,x)\) such that for \(f(x)=\sin x\),

\[\begin{aligned} \frac{f(x) ~-~ f(0)}{x - 0} ~=~ f'(c) \quad&\Rightarrow\quad \frac{\sin\,x ~-~ \sin\,0}{x - 0} ~=~ \cos\,c\

\[6pt] &\Rightarrow\quad \sin\,x ~=~ x\,\cos\,c\\ &\Rightarrow\quad \sin\,x ~\le~ x\end{aligned} \nonumber \]

since \(\cos\,c \le 1\) and \(x > 0\). Note that \(\sin\,x \le x\) is a sharper inequality than \(\sin\,x \le 1\) when \(0 < x < 1\).

There is a useful alternative form of the Mean Value Theorem. If \(a < b\) then let \(h=b-a>0\), so that \(a+h=b\). Then any number \(c\) in \((a,b)\) can be written as \(c = a + \theta h\) for some number \(\theta\) in \((0,1)\). To see this, let \(c\) be in \((a,b)\). Then \(0<c-a<b-a=h\) and so \(0<\frac{c-a}{h}<1\). Thus, \(\theta = \frac{c-a}{h}\) is in \((0,1)\) and \(a + \theta h = a + (c-a)=c\). Hence:

The Mean Value Theorem is the special case of \(g(x)=x\) in the following generalization:

The Mean Value Theorem says that the derivative of a differentiable function will always attain one particular value on a closed interval: the function’s average rate of change over the interval. It turns out that the derivative will take on every value between its values at the endpoints, similar to how the Intermediate Value Theorem applies to continuous functions:¹¹

In other words, if \(f'(a) < \gamma < f'(b)\) (or \(f'(b) < \gamma < f'(a)\)) then there is a number \(c\) in \((a,b)\) such that \(f'(c) = \gamma\). If \(f'\) were continuous on \(\ival{a}{b}\) then the result would follow trivially by the Intermediate Value Theorem for continuous functions. What is perhaps surprising is that Darboux’s Theorem holds even for derivatives that are not continuous. This means that a discontinuous derivative cannot have the type of simple jump discontinuities that would allow it to “skip” over intermediate values—the points of discontinuity must be of a more complicated type. One rough interpretation of Darboux’s Theorem is that even if a derivative is not a continuous function, it will behave sort of as if it were.

[sec4dot4]

Does Rolle’s Theorem apply to the function \(f(x) = 1 - \abs{x}\) on the interval \(\ival{-1}{1}\)? If so, find the number in \((-1,1)\) that Rolle’s Theorem guarantees to exist. If not, explain why not.

Suppose that two horses run a race starting together and ending in a tie. Show that, at some time during the race, they must have had the same speed.

Use the Mean Value Theorem to show that \(\Abs{\sin\;A ~-~ \sin\;B} ~\le~ \Abs{A-B}\) for all \(A\) and \(B\) (in radians). Does \(\Abs{\sin\;A ~+~ \sin\;B} ~\le~ \Abs{A+B}\) for all \(A\) and \(B\)? Explain.

Show that \(\Abs{\cos\;A ~-~ \cos\;B} ~\le~ \Abs{A-B}\) for all \(A\) and \(B\) (in radians). Does \(\Abs{\cos\;A ~+~ \cos\;B} ~\le~ \Abs{A+B}\) for all \(A\) and \(B\)? Explain.

Show that \(\tan\,x ~\ge~ x\) for all \(0 \le x < \frac{\pi}{2}\).

Show that \(\Abs{\tan\;A ~-~ \tan\;B} ~\ge~ \Abs{A-B}\) for all \(A\) and \(B\) (in radians) in \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). Can the inequality be extended to all \(A\) and \(B\)? Explain your answer. [[1.]]

Use Rolle’s Theorem to show that for all constants \(a\) and \(b\) with \(a>0\), \(f(x) = x^3 - ax + b\) can not have three positive roots. Also, show that it can not have three negative roots.

Use the Mean Value Theorem to show that if \(f'<0\) on an interval \(I\) then \(f\) is decreasing on \(I\).

Suppose that \(f\) and \(g\) are continuous on \(\ival{a}{b}\) and differentiable on \((a,b)\), and that \(f'(x) > g'(x)\) for all \(a < x < b\). Show that \(f(b) - g(b) > f(a) - g(a)\).

Prove the Extended Mean Value Theorem, by applying Rolle’s Theorem to the function

\[F(x) ~=~ f(x) ~-~ f(a) ~-~ \frac{f(b) - f(a)}{g(b) - g(a)}\,(g(x) - g(a)) ~. \nonumber \]

[exer:exple1px] Show that \(e^x \ge 1 + x\) for all \(x\). (Hint: Consider \(f(x)=e^x-x\).)

2

[exer:lnxltx] Show that \(\ln\,(1+x) < x\) for all \(x > 0\).

Show that \(\tan^{-1} x < x\) for all \(x\).

Show that for \(0 < \alpha \le \beta < \frac{\pi}{2}\),

\[\frac{\beta - \alpha}{\cos^2 \alpha} ~\le~ \tan\,\beta ~-~ \tan\,\alpha ~\le~ \frac{\beta - \alpha}{\cos^2 \beta} ~. \nonumber \]

Show that for \(0 < a \le b\),

\[\frac{b - a}{b} ~\le~ \ln\,\frac{b}{a} ~\le~ \frac{b - a}{a} ~. \nonumber \]

Show that for \(n > 1\) and \(a > b\),

\[nb^{n-1}(a-b) < a^n - b^n < na^{n-1}(a-b) ~. \nonumber \]

[[1.]]

Show that \(\sqrt{a^2 + b} \;<\; a + \dfrac{b}{2a}\) for all positive numbers \(a\) and \(b\).

Show that \(f(x) \;=\; \cos^2 x \;+\; \cos^2\left(\frac{\pi}{3}+x\right) \;-\; \cos\,x\, \cos\,\left(\frac{\pi}{3}+x\right)\) is a constant function. What is its value?

Suppose that \(f(x)\) is a differentiable function and that \(f(0)=0\) and \(f(1)=1\). Show that \(f'(x_0) = 2x_0\) for some \(x_0\) in the interval \((0,1)\).

Prove the inequality

\[\ABS{\frac{x_1 + x_2}{1 + x_1 x_2}} ~<~ 1 \quad\text{for $\;-1 ~<~ x_1, x_2 ~<~ 1$} \nonumber \]

as follows:

1. First prove the special case where \(x_1 = x_2\).
2. For the case \(x_1 < x_2\) define

   \[f(x) = \frac{x+a}{1+ax} \nonumber \]

   for \(-1 \le x \le 1\), where \(-1 < a < 1\). Show that \(f\) is increasing on \(\ival{-1}{1}\), then use \(a=x_2\) and \(x=x_1\).

Note that proving the case \(x_2 < x_1\) is unnecessary (why?).
This inequality is a generalization of the same inequality for \(0 \le x_1, x_2 < 1\) in the relativistic velocity addition law from the theory of special relativity: if object 1 has velocity \(v_1\) relative to a frame of reference \(F\), and if object 2 has a velocity \(v_2\) relative to object 1, so that \(x_1 = v_1/c\) and \(x_2 = v_2/c\) represent the fractions of the speed of light \(c\) at which the objects are moving, then the fraction of the speed of light at which object 2 is moving with respect to \(F\) is \(x = (x_1 + x_2)/(1 + x_1 x_2)\). So it should be true that \(0 \le x < 1\), since nothing can move faster than the speed of light.