3.8  PROPERTIES  OF  CONTINUOUS  FUNCTIONS 

This section develops some theory that will be needed for integration in Chapter 4. We begin with a new concept, that of a hyperinteger. The hyperintegers are to the integers as the hyperreal numbers are to the real numbers. The hyperintegers consist of the ordinary finite integers, the positive infinite hyperintegers, and the negative infinite hyperintegers. The hyperintegers have the same algebraic properties as the integers and are spaced one apart all along the hyperreal line as in Figure 3.8.1.

 

 

 

 

 

 

 

 

 

 

Figure  3.8.1   The Set of Hyperintegers

 

The rigorous definition of the hyperintegers uses the greatest integer function [x] introduced in Section 3.4, Example 6. Remember that for a real number x, [x] is the greatest integer n such that n ≤ x. A real number y is itself an integer if and only if y= [x] for some real x. To get the hyperintegers, we apply the function [x] to hyperreal number x (see Figure 3.8.2).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 3.8.2

 

DEFINITION

 

      A hyperinteger is a hyperreal number y such that y=[x] for some hyperreal x.

 

      When x varies over the hyperreal numbers, [x] is the greatest hyperinteger y such that y ≤ x. Because of the Transfer Principle, every hyperreal number x is between two hyperintegers [x] and [x] + 1,

                              [x] ≤ x ≤ [x]+1.

Also, sums, differences, and products of hyperintegers are again hyperintegers.

 

  We are now going to use the hyperintegers. In sketching curves we divided a closed interval [a, b] into finitely many subintervals. For theoretical purposes in the calculus we often divide a closed interval into a finite or infinite number of equal subintervals. This is done as follows.

 

  Given a closed real interval [a, b], a finite partition is formed by choosing a positive integer n and dividing [a, b] into n equal parts, as in Figure 3.8.3. Each part will be a subinterval of length t = (b-a)/n. The n subintervals are

                  [a, a+ t ], [a+ t, a+ 2t], ……, [a+ (n -1) t, b].

 

 

 

 

Figure  3.8.3

 

The endpoints

                        a, a+ t, a+ 2t, ……, a+ (n-1) t, a+nt = b

 

are called partition points.

 

The real interval [a,b] is contained in the hyperreal interval [a,b] *, which is the set of all hyperreal numbers x such that a ≤ x ≤ b. An infinite partition is applied to the hyperreal interval [a,b] * rather than the real interval. To form an infinite partition of [a,b]*, choose a positive infinite hyperinteger H and divide [a,b]* into H equal parts as shown in Figure 3.8.4. Each subinterval will have the same infinitesimal length δ =(b-a)/H. The H subintervals are

 

                  [ a, a+δ ] , [ a+δ, a+2δ], ……[a+(K-1)δ, a+Kδ], ……[a+(H -1)δ, b],

and the partition points are

                   a, a+δ, a+2δ, ……a+Kδ，……a+Hδ = b,

Where K runs over the hyperintegers from 1 to H. Every hyperreal number x between a and b belongs to one of the infinitesimal subintervals,

                  a+(K-1)δ ≤ x< a+Kδ.

 

 

 

                

 

 

 

 

 

 

 

 

 

We shall now use infinite partitions to sketch the proofs of three basic results, called the Intermediate Value Theorem, the Extreme Value Theorem, and Rolle’s Theorem. The use of these results will be illustrated by studying zeros of continuous functions. By a zero of a function f we mean a point c where f(c) = 0. As we can see in Figure 3.8.5, the zeros of f are the points where the curve y=f(x) intersects the x-axis.

 

 

 

 

 

 

 

 

 

 

Figure  3.8.5

 

INTERMEDIATE  VALUE  THEOREM

 

        Suppose the real function f is continuous on the closed interval [a,b] and f(x) is positive at one

        endpoint and negative at the other endpoint. Then f has a zero in the interval (a,b) ; that is, f(c)=0

        for some real c in (a,b).

 

Discussion There are two cases illustrated in Figure 3.8.6:

                       f(a) <0 0 > f(b).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure  3.8.6

 

In the first case, the theorem says that if a continuous curve is below the x-axis at a and above it at b, then the curve must intersect the x-axis at some point c between a and b. Theorem 3 in the preceding Section 3.7 on curve sketching is simply a reformulation of the Intermediate Value Theorem.

 

SKETCH  OF  PROOF   We assume f(a) ≤ 0 ≤ f(b). Let H be a positive infinite hyperinteger and partition

     the interval [a,b]* into H equal parts

 

                       a, a+ δ, a+ 2δ,……a+Hδ = b.

Let a+Kδ be the last partition point at which f(a+Kδ) < 0. Thus

                       f(a+Kδ) < 0 ≤ f(a+ (K+1)δ).

Since f is continuous, f(a+Kδ) is infinitely close to f(a+(K+1)δ). We conclude that f(a+ Kδ) ≈ 0 (Figure 3.8.7). We take c to be the standard part of a+ Kδ, so that

                       f(c) = st (f (a + Kδ)) = 0.

 

EXAMPLE 1  The function

 

 

 

            Which is shown in Figure 3.8.8, is continuous for 0≤ x ≤ 1. Moreover,

                     

                            f(0) = 1,

 

            The Intermediate Value Theorem shows that f(x) has a zero f(c) = 0 for some c between 0 and 1.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 3.8.7

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure  3.8.8

 

The Intermediate Value Theorem can be used to prove Theorem 3 of Section 3.7 on curve sketching:

 

Suppose g is a continuous function on an interval I, and g(x) ≠ 0 for all x in I.

 

(i)  If g(c) > 0 for at least one c in I, then g(x) >0 for all x in I.

(ii)  If g(c)<0 for at least one c in I, then g(x) <0 for all x in I.

 

PROOF (i)  Let g(c) > 0 for some c in I. If g(x1) < 0 for some other point x1 in I, then by the Intermediate Value

         Theorem there is a point x2 between c and x1 such that g(x2) = 0, contrary to hypothesis (Figure 3.8.9).

         Therefore we conclude that g(x) >0 for all x in I.

 

 

Figure  3.8.9

 

EXTREME  VALUE  THEOREM

 

Let f be continuous on its domain, which is a closed interval [a, b]. Then f has a maximum at some point in [a, b], and a minimum at some point in [a, b].

 

Discussion 

We have seen several examples of functions that do not have maxima on an open interval, such as f(x) = 1/x

on ( 0, ∞), or g(x) = 2x on (0, 1). The Extreme Value Theorem says that on a closed interval a continuous function always has maximum.

 

SKETCH  OF  PROOF     Form an infinite partition of [a, b]*,

                                  a, a+δ, a+2δ,……，a+Hδ=b

        By the Transfer Principle, there is a partition point a+ Kδ at which f(a+ Kδ ) has the largest value.

        Let c be the standard part of a + Kδ (see Figure 3.8.10). Any point u of [a, b]* lies in a subinterval, say

                                  a, Lδ, ≤ u < a+(L+1)δ.

 

        We have                      f(a + Kδ) ≥ f(a + Lδ)

        and taking standard parts,

                                    f(c) ≥ f(u)

 

This shows that f has a maximum at c.

 

 

 

 

 

 

 

 

 

 

 

 

Figure 3.8.10

 

ROLLE’ S  THEOREM

Suppose that f is continuous on the closed interval [ a, b] and differentiable on the open interval (a, b). If

                                 f(a) = f(b) = 0,

 

then there is at least one point c strictly between a and b where f has derivative zero; i. e.,

                 f ′ (c) = 0         for some c in (a, b).

Geometrically, the theorem says that a differentiable curve touching the x-axis at a and b must be horizontal for at least one point strictly between a and b.

 

Proof   We may assume that [a, b] is the domain of f. By the Extreme Value Theorem, f has a maximum value

        M and a minimum value m in [a, b]. Since f(a) = 0, m ≤ 0 and M ≥ 0 (see Figure 3.8.11).

 

Case 1  M = 0 m = 0. Then f is the constant function f(x) = 0, and therefore f ′(c) = 0 for all points c in (a, b).

Case 2 M > 0. Let f have a maximum at c, f(c) = M. By the Critical Point Theorem, f has a critical point at c . c

        cannot be an endpoint because the value of f(x) is zero at the endpoints and positive at x= c.

        By hypothesis, f ′(x) exists at x=c. It follows that c must be a critical point of the type f ′(c) = 0.

 

Case 3  m < 0. We let f have a minimum at c. Then as in Case 2, c is in (a, b) and f ′(c) = 0.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure  3.8.11  Rolle’s Theorem

 

EXAMPLE  2

f(x) = (x-1)² (x-2) ³ , a= 1, b=2. The function f is continuous and differentiable everywhere (Figure 3.8.12). Moreover, f(1) = f(2) =0. Therefore by Rolle’s Theorem there is a point c in (1, 2) with f ′(c) = 0.

 

    Let us find such a point c. We have

       f ′(x) = 3(x-1)² (x-2)²  + 2(x-1) (x-2) ³ = (x -1) (x-2)² (5x-7).

 

 

 

 

 

 

 

 

 

 

 

 

Figure 3.8.12

 

Notice that f ′(1) = 0 and f ′(2) = 0. But Rolle’s Theorem says that there is another point c which is in the open interval (1,2) where f ′(c) = 0. The required value for c is c = ___ becasuse

 

EXAMPLE  3  Let 

 

         Then f(a) = f(b) = 0.

         Rolle’s Theorem says that there is at least one point c in             at which f ′(c) = 0.

         As a matter of fact there are three such points,

                                 c = -1,   c = 0,    c =1.

         We can find these points as follows:

                       f ′(x) = 2x³ - 2x = 2x(x²-1),

                       f ′(x) = 0   when  x= 0  or x= + 1.

 

         Then function is drawn in Figure 3.8.13.

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure  3.8.13

 

EXAMPLE 4     

 a= -1, b=1. Then f(-1) = f(1) = 0. Then function f is continuous on [-1,1] and has a derivative at each point of (-1,1), as Rolle’s Theorem requires (Figure 3.8.14). Note, however, that f ′(x) does not exist at either endpoint, x = -1 or x=1. By Rolle’s Theorem there is a point c in (-1,1) such that f ′(c) = 0, c = 0 is such a point, because

              

 

 

EXAMPLE 5

f(x)= 1- x2/3,  a= -1, b=1. Then f(-1) = f(1) =0, and f ′(x) = _______ for x ≠ 0. f ′(0) is undefined. There is no point c in (-1,1) at which f ′(c) = 0. Rolle’s Theorem does not apply in this case because f ′(x) does not exist at one of the points of the interval (-1,1), namely at x=0. In Figure 3.8.15, we see that instead of being horizontal at a point in the interval, the curve has a sharp peak.

 

Rolle’s Theorem is useful in finding the number of zeros of a differentiable function f. It shows that between any two zeros of f there must be one or more zeros of f ′. It follows that if f ′ has no zeros in an interval I , then f cannot have more than one zero in I.

 

EXAMPLE  6  

How many zeros does the function f(x) = x³ + x +1 have ? We use both Rolle’s Theorem and the Intermediate Value Theorem.

Using Rolle’s Theorem: f ′(x) = 3x² + 1. For all x, x² ≥ 0, and hence f ′(x) ≥ 1. Therefore f(x) has at most one zero.

Using the intermediate value theorem: we have f(-1) = -1, f(0) =1.

Therefore f has at least one zero between -1 and 0.

 

CONCLUSION  f has exactly one zero, and it lies between -1 and 0 ( see Figure 3.8.16).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 3.8.16

 

 

Our method of sketching curves in Section 3.7 depends on a consequence of Rolle’s Theorem called the Mean Value Theorem. It deals with the average slope of a curve between two points.

 

DEFINITION

 

   Let f be defined on the closed interval [a,b]. The average slope of f between a and b is the quotient

 

                       

We can see in Figure 3.8.17 that the average slope of f between a and b is equal to the slope of the line passing through the points (a, f(a)) and (b, f(b)). This is shown by the two-point equation for a line (Section 1.3). In particular, if f is already a linear function f(x) = mx + c, then the average slope of f between a and b is equal to the slope m of the line y=f(x).

 

 

 

 

 

 

 

 

 

Figure 3.8.17   Average  Slope

 

This is shown by the two-point equation for a straight line (Section 1.2). In particular, if f is already a linear function f(x) = mx + c, then the average slope of f between a and b is equal to the slope m of the straight line

 y= f(x).

 

MEAN  VALUE  THEOREM

 

        Assume that f is continuous on the closed interval [a, b] and has a derivative at every point of the open

        interval (a,b). Then there is at least one point c in (a,b) where the slope f(c) is equal to the average

        slope of f between a and b,

 

 

 

 

Remark  In the special case that f(a) = f(b) = 0, the Mean Value Theorem becomes Rolle’s Theorem:

                 

 

 

On the other hand, we shall use Rolle’s Theorem in the proof of the Mean Value Theorem. The Mean Value Theorem is illustrated in Figure 3.8.18.

 

 

 

 

 

 

 

 

 

 

 Figure 3.8.18  the Mean Value Theorem

 

PROOF OF THE MEAN VALUE THEOREM

Let m be the average slope, m= ( f(b) - f(a)) / (b-a). The line through the points (a, f(a)) and (b, f(b)) has the equation

                       ι (x) = f(a) + m(x-a).

Let h(x) be the distance of f(x) above I(x),

                          h(x) = f(x) - ι(x).

Then h is continuous on [a,b] and has the derivative

                          h′(x) = f ′(x) - ι′(x) = f ′(x) - m

at each point in (a,b). Since f(x) = ι(x) at the endpoints a and b, we have

                          h(a) = 0,  h(b) = 0.

Therefore Rolle’s Theorem can be applied to the function h, and there is a point c in (a, b) such that h′(c) = 0. Thus

                           0= h′(c) = f ′(c) - ι ′(c) = f ′(c) - m,

Whence                     f ′(c) = m.

 

We can give a physical interpretation of the Mean Value Theorem in terms of velocity. Suppose a particle moves along the y-axis according to the equation y=f(t). The average velocity of the particle between times a and b is the ratio

      

 

        

 

of the change in position to the time elapsed. The Mean Value Theorem states that there is a point of time c, a, when the velocity f ′(c) of the particle is equal to the average velocity between times a and b.

 

Theorems 1 and 2 in Section 3.7 on curve sketching are consequences of the Mean Value Theorem. As an illustration, we prove part (ii) of Theorem1:

           If f ′(x) > 0 for all interior points x of I, then f is increasing on I.

          

 

PROOF  Let x12 where x1 and x2 are points in I. By the Mean Value Theorem there is a point c strictly

         between x1 and x2 such that

 

 

 

 

 

Since c is an interior point of I, f ′(c) > 0. Because x1<<i>x2,  x2-x1 > 0 .

Thus

 

 

This shows that f is increasing on I.

 

 

PROBLEMS  FOR  SECTION  3.8

In Problems 1-16, use the Intermediate Value Theorem to show that the function has at least one zero in the given interval.

1    f(x) = x4 - 2x³ - x² + 1,  0 ≤ x ≤1

2    f(x) = x² +x - 3/x ,  1 ≤ x ≤ 2

 

3    

 

4

5

 

6   ______________________________

7   f(x)=x³ + x² -1,  0 ≤ x ≤ 1

8

 

 

9    f(x) = 1-3x+x³,  0 ≤ x ≤ 1

 

10    f(x) = 1-3x+x³,  1 ≤ x ≤ 2

11    ___________________

12    f(x) = x² -(x+1)-1/2,  0 ≤ x ≤ 1

13    ________________

14    f(x) = sin x - 2cos x, 0 ≤ x ≤ π

15   ____________________

16   f(x) = ex -10x , 1 ≤ x ≤ 10

 

 

 

In Problems 17-30, determine whether or not f ′ has a zero in the interval (a, b). Warning: Rolle’s Theorem may give a wrong answer unless all the hypotheses are met.

 

17    ____________________

18    f(x)=1 – x -2,  [a,b] = [ -1, 1]

19   _____________________

20   ___________________

21   f(x)=1/x – x,  [a,b] = [ -1, 1]

22   f(x)=(x -1) 2 ( x-2),  [a,b] = [ 1, 2]

23   f(x)=(x -4) 3 x4,  [a,b] = [ 0, 4]

24   __________________________

25   f(x)= |x |– 1,  [a,b] = [ -1, 1]

26   _____________________

27   f(x) = x sin x, [a,b] = [0, π]

28   f(x) = ex cos x, [a,b] = [-π/2, π/2]

 

29    f(x) = tan x, [a, b] = [0, π]

30    f(x) = 1n (1-sin x), [a, b] = [0, π]

31    Find the number of zeros of x4+3x+1 in [-2, -1].

32    Find the number of zeros of x4+2 x3-2 in [0, 1].

33   Find the number of zeros of x4-8x-4.

34    Find the number of zeros of

 

 

In Problems 35-42, find a point c in (a, b) such that f(b) -f(a) =f ′(c) (b-a).

35    f(x) = x2 +2x – 1,  [a, b]= [0, 1]

36    f(x) = x3 ,  [a, b]= [0, 3]

37    f(x) = x2/3 ,  [a, b]= [0, 1]

38   ________________

39   _______________

40   f(x) = 2 + (1/x),  [a, b]= [1, 2]

41

 

42   

 

43  Use Rolle’s Theorem to show that the function f(x) = x3 - 3x + b cannot have more than

one zero in the interval [ -1,1 ], regardless of the value of the constant b.

44  Suppose f, f ′, and f ′′ are all continuous on the interval [a, b], and suppose f has at

least three distinct zeros in[a, b]. Use Rolle’s Theorem to show that f ′′ has at least one

zero in [a, b].

45 Suppose that f ′′(x) > 0 for all real numbers x, so that the curve y= f(x) is concave

upward on the whole real line as illustrated in the figure. Let L be the tangent line to the

curve at x= c. Prove that the line L lies below the curve at every point x ≠ c

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

EXTRA  PROBLEMS  FOR  CHAPTER  3

1   Find the surface area A of a cube as a function of its volume V.

2   Find the length of the diagonal d of a rectangle as a function of its length x and width y.

3   An airplane travels for t hours at a speed of 300 mph.

Find the distance x of travel as a function of t.

4   An airplane travels x miles at 500 mph. Find the travelling time t as a function of x.

5   A 5 foot tall woman stands at a distance x from a 9 foot high lamp. Find the length of

her shadow as a function of x.

6   The sides and bottom of a rectangular box are made of material costing $1/sq ft. and the

top of material costing $2/sq ft. Find the cost of the box as a function of the length x,

width  y, and height z feet.

7   A piece of dough with a constant volume of 10 cu in. is being rolled in the shape of a right circular cylinder. Find the rate of increase of its length when the radius is ___ inch and is decreasing at __ inch per second.

8   Car A travels north at 60 mph and passes the point P at 1:00. Car B travels east at 40

mph and passes the point P at 3:00. Find the rate of change of the distance between the

two cars at 2:00.

10  A country has a constant national income and its population is decreasing by one million people per year. Find the rate of change of the per capita income when the population is 50 million and the national income is 100 billion dollars.

 

 

 

 

 

 

 

 

 

15Find the set of all points at which                    

is continuous.

16Find the set of all points at which

              

 

 

is continuous.

17Find the set of all points at which is continuous.

18Assume a<b. Show that is continuous on the closed interval [a,b].

19Show that g(x) = (x-1) 1/3 is continuous at every real number x=c.

20Find the maximum and minimum of

                            f(x) = 4x3-3x2 + 2,    -1≤ x ≤1.

21Find the maximum and minimum of

                      _________________________

22Find the maximum and minimum of

                  f(x)= |2x-5|+3,   0 ≤ x ≤10.

23Find the maximum and minimum of

                 f(x)= 4-3x2/3,   -1 ≤ x ≤1.

24Find the maximum and minimum of

f(x)= (x-1)1/3-2,   0≤ x ≤2.

25find the rectangle of maximum area which can be inscribed in a circle of radius 1.

26A box with a square base and no top is to be made with 10 sq ft of material. Find the dimensions which will have the largest volume.

27In one day a factory can produce x items at a total cost of c0+ax dollars and can sell x items at a price of bx-1/3 dollars per item. How many items should be produced for a maximum daily profit?

28Test the curve f(x) = x3 -5x+4 for maxima and minima.

29Test the curve f(x) = 3x4 +4 x3-12 x2 for maxima and minima.

30The light intensity from a light source is equal to S/D2 where S is the strength of the source and D the distance from the source. Two light sources A and B have strengths SA=2 and SB=1 and are located on the x-axis at xA=0 and xB=10. Find the point x, 0< x <10, where the total light intensity is a minimum.

31Find the right triangle of area ___ with the smallest perimeter.

32Find the points on the parabola y=x2 which are closest to the point (0, 2).

33Find the number of zeros of f(x) = x3-8x2+4x+2.

34Find the number of zeros of f(x) = x3-2x2+2x-4.

35Sketch the curve y= x4-x3,   -1≤x≤1

36Sketch the curve y= x2+x-2,   ____≤x≤2.

37Find all zeros of f(x) = x2 -5x+10.

38Show that the function f(x) = x6 -5x5- 3x2+4 has at least one zero in the interval [0, 1].

39Show that the function ________________ has at least one zero in the interval [-1, 0].

40Show that the equation ________________ has at least one solution in the interval [0,1].

41  Prove that _________________ exists if and only if there is a function g(x) such that

(a)g(x) is continuous at x=c.

(b)g(x) =f(x) whenever x≠c.

42  Let S = {a1,…an} be a finite set of real numbers. Show that the characteristic function of S,

                      

 

 

 

     is discontinuous for x in S and continuous for x not in S.

43  Show that the function f(x)=____ is continuous but not differentiable at x=0.

44  Let

                      

 

 

 

     Show that f is continuous at x=0 but discontinuous at x = 1/n and x= -1/n, n = 1,2,3,…

45  Let

    

 

 

 

 

     Prove that f is differentiable at x=0 but discontinuous at x=1/n and x= -1/n, n=1,2,3,…

46  Suppose f(x) is continuous on [0,1] and f(0)=1, f(1) =0. Prove that there is a point c in (0, 1)

such that f(c ) = c.

47  Suppose f(x) is continuous for all x, and f(0)=0, f(1) =4, f(2) =0. Prove that there is a point c

in (0, 1) such that f(c ) = f(c+1).

48  Prove that if x=c is the only real solution of f(x)= 0, then x=c is also the only hyperreal

solution.

49  Prove that if n is odd, then the polynomial

                      xn+an-1 x n-1 + …a1x+a0

      has no maximum and no minimum.

50  Prove that if n is even then the polynomial

xn+an-1 x n-1 + …a1x+a0

has no maximum.

 

51  Prove that if n is even then the polynomial

xn+an-1 x n-1 + …a1x+a0

     has a minimum. You may use the fact that there are only finitely many critical points.

52  Prove the First Derivative Test: Assume f(x) is continuous on an interval I.

     If f ′(a) > 0 for all a < c and f ′(b) <0 for all b> c , then f has a maximum at x=c.

     If f ′(a) < 0 for all a < c and f ′(b) >0 for all b> c , then f has a minimum at x=c.

53 Suppose f is differentiable and f ′(x) >1 for all x. If f(0)=0, show that f(x) > x for all positive x.

54 Suppose f ′′>0 for all x. Show that for any two points P and Q above the curve y=f(x), every

point on the line segment PQ is above the curve y=f(x).

55 Suppose f (0)=A and f ′(x) has the constant value B for all x. Use the Mean Value Theorem to

show that f is the linear function f(x) = A + Bx.

56 Suppose f′′ (x) is continuous for all real x. Use the Mean Value Theorem to show that for all

finite hyperreal b and nonzero infinitesimal Δ x,

 

 

 

 

 

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