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4.2: THE MEAN VALUE THEOREM

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In this section, we focus on the Mean Value Theorem, one of the most important tools of calculus and one of the most beautiful results of mathematical analysis. The Mean Value Theorem we study in this section was stated by the French mathematician Augustin Louis Cauchy (1789-1857), which follows form a simpler version called Rolle's Theorem.

An important application of differentiation is solving optimization problems. A simple method for identifying local extrema of a function was found by the French mathematician Pierre de Fermat (1601-1665). Fermat's method can also be used to prove Rolle's Theorem.

We start with some basic definitions of minima and maxima. Recall that for aR and δ>0, the sets B(a;δ), B+(a;δ), and B(a;δ) denote the intervals (aδ,a+δ), (a,a+δ) and (aδ,a), respectively.

Definition 4.2.1

Let D be a nonempty subset of R and let f:DR. We say that f has a local (or relative) minimum at aD if there exists δ>0 such that

f(x)f(a) for all xB(a;δ)D.

Similarly, we say that f has a local (or relative) maximum at aD if there exists δ>0 such that

f(x)f(a) for all xB(a;δ)D.

In January 1638, Pierre de Fermat described his method for finding maxima and minima in a letter written to Marin Mersenne (1588-1648) who was considered as "the center of the world of science and mathematics during the first half of the 1600s." His method presented in the theorem below is now known as Fermat's Rule.

Theorem 4.2.1 - Fermat's Rule.

Let I be an open interval and f:IR. If f has a local minimum or maximum at aI and f is differentiable at a, then f(a)=0.

clipboard_e93529c349b5b1d5eaa7ef48bfa1c1985.png

Figure 4.1: Illustration of Fermat's Rule.

Proof

Suppose f has a local minimum at a. Then there exists δ>0 sufficiently small such that

f(x)f(a) for all xB(a;δ).

Since B+(a;δ) is a subset of B(a;δ), we have

f(x)f(a)xa0 for all xB+(a;δ).

Taking into account the differentiability of f at a yields

f(a)=limxaf(x)f(a)xa=limxa+f(x)f(a)xa0.

Similarly,

f(x)f(a)xa0 for all xB(a;δ).

It follows that

f(a)=limxaf(x)f(a)xa=limxaf(x)f(a)xa0.

Therefore, f(a)=0. The proof is similar for the case where f has a local maximum at a.

Theorem 4.2.2 - Rolle's Theorem.

Let a,bR with a<b and f:[a,b]R. Suppose f is continuous on [a,b] and differentiable on (a,b) with f(a)=f(b). Then there exists c(a,b) such that

f(c)=0.

Proof

Since f is continuous on the compact [a,b], by the extreme value theorem (Theorem 3.4.2) there exists ˉx1[a,b] and ˉx2[a,b] such that

f(ˉx1)=min{f(x):x[a,b]} and f(ˉx2)=max{f(x):x[a,b]}.

Then

f(ˉx1)f(x)f(ˉx2) for all x[a,b].

Annotation 2020-09-01 205639.png

Figure 4.2: Illustration of Rolle's Theorem.

If ˉx1(a,b) or ˉx2(a,b), then f has a local minimum at ˉx1 or f has a local maximum at ˉx2. By Theorem 4.2.1, f(ˉx1)=0 or f(ˉx2)=0, and (4.3) holds with c=ˉx1 or c=ˉx2.

If both ˉx1 and ˉx2 are the endpoints of [a,b], then f(ˉx1)=f(ˉx2) because f(a)=f(b). By (4.4), f is a constant function, so f(c)=0 for any c(a,b).

We are now ready to use Rolle's Theorem to prove the Mean Value Theorem presented below.

clipboard_e2d72657b7ec5a4e91f7bbbbb494e9668.png

Figure 4.3: Illustration of the Mean Value Theorem.

Theorem 4.2.3 - Mean Value Theorem.

Let a,bR with a<b and f:[a,b]R. Suppose f is continuous on [a,b] and differentiable on (a,b). Then there exists c(a,b) such that

f(c)=f(b)f(a)ba.

Proof

The linear function whose graph goes through (a,f(a)) and (b,f(b)) is

g(x)=f(b)f(a)ba(xa)+f(a).

Define

h(x)=f(x)g(x)=f(x)[f(b)f(a)ba(xa)+f(a)] for x[a,b].

Then h(a)=h(b), and h satisfies the assumptions of Theorem 4.2.2. Thus, there exists c(a,b) such that h(c)=0. Since

h(x)=f(x)f(b)f(a)ba,

it follows that

f(c)f(b)f(a)ba=0.

Thus, (4.5) holds.

Example 4.2.1

We show that |sinx||x| for all xR.

Solution

Let f(x)=sinx for all xR. Then f(x)=cosx. Now, fix xR, x>0. By the Mean value Theorem applied to f on the interval [0,x], there exists c(0,x) such that

sinxsin0x0=cosc.

Therefore, |sinx||x|=|cosc|. Since |cosc|1 we conclude |sinx||x| for all x>0. Next suppose x<0. Another application of the Mean Value Theorem shows there exists c(x,0) such that

sin0sinx0x=cosc.

Then, again, |sinx||x|=|cosc|1. It follows that |sinx||x| for x<0. Since equality holds for x=0, we conclude that |sinx||x| for all xR.

Example 4.2.2

We show that 1+4x<(5+2x)/3 for all x>2.

Solution

Let f(x)=1+4x for all x2. Then

f(x)=421+4x=21+4x.

Now, fix xR such that x>2. We apply the Mean Value Theorem to f on the interval [2,x]. Then, since f(2)=3, there exists c(2,x) such that

1+4x3=f(c)(x2).

Since f(2)=2/3 and f(c)<f(2) for c>2 we conclude that

1+4x3<23(x2).

Rearranging terms provides the desired inequality.

A more general result which follows directly from the Mean Value Theorem is known as Cauchy's Theorem.

Theorem 4.2.4 - Cauchy's Theorem.

Let a,bR with a<b. Suppose f and g are continuous on [a,b] and differentiable on (a,b). Then there exists c(a,b) such that

[f(b)f(a)]g(c)=[g(b)g(a)]f(c).

Proof

Define

h(x)=[f(b)f(a)]g(x)[g(b)g(a)]f(x) for x[a,b].

Then h(a)=f(b)g(a)f(a)g(b)=h(b), and h satisfies the assumptions of Theorem 4.2.2. Thus, there exists c(a,b) such that h(c)=0. Since

h(x)=[f(b)f(a)]g(x)[g(b)g(a)]f(x),

this implies (4.6).

The following theorem shows that the derivative of a differentiable function on [a,b] satisfies the intermediate value property although the derivative function is not assumed to be continuous. To give the theorem in its greatest generality, we introduce a couple of definitions.

Definition 4.2.2

Let a,bR, a<b, and f:[a,b]R. If the limit

limxa+f(x)f(a)xa

exists, we say that f has a right derivative at a and write

f+(a)=limxa+f(x)f(a)xa.

If the limit

limxbf(x)f(b)xb

exists, we say that f has a left derivative at b and write

f(b)=limxbf(x)f(b)xb.

We will say that f is differentiable on [a,b] if f(x) exists for each x(a,b) and, in addition, both f+(a) and f(b) exist.

Theorem 4.2.5 - Intermediate Value Theorem for Derivatives.

Let a,bR with a<b. Suppose f is differentiable on [a,b] and

f+(a)<λ<f(b).

Then there exists c(a,b) such that

f(c)=λ.

clipboard_edce8aa985685b55e13573ce64e8efad2.png

Figure 4.4: Right derivative.

Proof

Define the function g:[a,b]R by

g(x)=f(x)λx.

Then g is differentiable on [a,b] and

g+(a)<0<g(b).

Thus,

limxa+g(x)g(a)xa<0.

It follows that there exists δ1>0 such that

g(x)<g(a) for all x(a,a+δ1)[a,b].

SImilarly, there exists δ2>0 such that

g(x)<g(b) for all x(bδ2,b)[a,b].

Since g is continuous on [a,b], it attains its minimum at a point c[a,b]. From the observations above, it follows that c(a,b). This implies g(c)=0 or, equivalently, that f(c)=λ. square

Remark 4.2.6

The same conclusion follows if f+(a)>λ>f(b).

Exercise 4.2.1

Let f and g be differentiable at x0. Suppose and

f(x)g(x) for all xR.

Prove that f(x0)=g(x0).

Answer

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Exercise 4.2.2

Prove the following inequalities using the Mean Value Theorem.

  1. 1+x<1+12x for x>0.
  2. ex>1+x, for x>0. (Assume known that the derivative of ex is itself.)
  3. x1x<lnx<x1, for x>1. (Assume known that the derivative of lnx is 1/x.)
Answer

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Exercise 4.2.3

Prove that |sin(x)sin(y)||xy| for all x,yR.

Answer

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Exercise 4.2.4

Let n be a positive integer and let ak,bkR for k=1,,n. Prove that the equation

x+nk=1(aksinkx+bkcoskx)=0

has a solution on (π,π).

Answer

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Exercise 4.2.5

Let f and g be differentiable functions on [a,b]. Suppose g(x)0 and g(x)0 for all x[a,b]. Prove that there exists c(a,b) such that

1g(b)g(a)f(a)f(b)g(a)g(b)∣=1g(c)f(c)g(c)f(c)g(c),

where the bars denote determinants of the two-by-two matrices.

Answer

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Exercise 4.2.6

Let n be a fixed positive integer.

  1. Suppose a1,a2,,an satisfy

a1+a22++ann=0.

Prove that the equation

a1+a2x+a3x2++anxn1=0

has a solution in (0,1).

  1. Suppose a0,a1,,an satisfy

nk=0ak2k+1=0.

Prove that the equation

nk=0akcos(2k+1)x=0

has a solution on (0,π2).

Answer

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Exercise 4.2.7

Let f:[0,)R be a differentiable function. Prove that if both limxf(x) and limxf(x) exist, then limxf(x)=0

Answer

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Exercise 4.2.8

Let f:[0,)R be a differentiable function.

  1. Show that if limxf(x)=a, then limxf(x)x=a.
  2. Show that if limxf(x)=, then limxf(x)x=.
  3. Are the converses in part (a) and part (b) true?
Answer

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This page titled 4.2: THE MEAN VALUE THEOREM is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Lafferriere, Lafferriere, and Nguyen (PDXOpen: Open Educational Resources) .

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