Intermediate Value Theorem

7 Key excerpts on "Intermediate Value Theorem"

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  eBook - ePub

  Introduction to Differential Calculus

  Systematic Studies with Engineering Applications for Beginners

  • Ulrich L. Rohde, G. C. Jain, Ajay K. Poddar, A. K. Ghosh(Authors)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  Existence Theorems , the MVT being one of them. Some other examples of existence theorems are the Intermediate Value Theorem (IVT) and Extreme Value Theorem (EVT), both pertaining to continuous functions defined on closed and bounded intervals. Without going into the proof of these theorems, we indicate why the property each guarantees is practically useful.

  I. Intermediate Value Theorem [Already introduced in Chapter 8, but again repeated here for ready reference]: Let f be continuous on the closed and bounded interval [a , b ] and let y be any number between f (a ) and f (b ). Then, there exists a number c between a and b for which f(c) = y .

  The IVT says that, if f is continuous on [a , b ], then the range of f contains not just f(a) and f(b) but everything in between . This means that the graph of a continuous function f is unbroken. In other words, it means that enroute from (a , f (a )) to (b , f (b )), the graph of f crosses every horizontal line at one (or more number of points) between y = f (a ) and y = f (b ).

  II. Extreme Value Theorem : Let f be continuous on the closed and bounded interval [a , b ]. Then, f assumes both a maximum value and a minimum value somewhere on [a , b ].

  The EVT guarantees that, if f is continuous on the closed interval [a , b ], then f attains both a maximum and a minimum somewhere therein.

  [Both hypotheses of the EVT—that f be continuous and that the interval be closed are necessary. If either fails, f need not assume a maximum or a minimum.]

  Note (8): Each of these existence theorems guarantees the existence of at least one point in the domain with some desirable property. Neither theorem states where in [a , b

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  Foundations of Mathematics

  Algebra, Geometry, Trigonometry and Calculus

  • Philip Brown(Author)
  • 2016(Publication Date)
  • Mercury Learning and Information

    (Publisher)

  7.7APPLICATIONS OF CONTINUITY

   

  Continuous functions are important because of their mathematical properties, which we will begin to explore in this chapter. We have already seen a connection between continuity and the evaluation of limits.

  Many processes in the real world happen in a continuous way (without breaks and jumps), and so these processes can be represented by continuous functions. For example, if a hiker walks from the bottom to the top of a mountain, then his altitude is a continuous function of the time elapsed. His altitude as a function of the horizontal distance he has walked could also be a continuous function (this function would not be continuous if the hiker has to climb up a vertical rope at a cliff face!).

  7.7.1The Intermediate Value Theorem

  We now state an important theorem relating to continuous real-valued functions:

  Theorem 7.7.1. The Intermediate Value Theorem: if a real-valued function f(x) is continuous on a closed interval [a, b] and v is any real number between f(a) and f(b) (assuming f(a) ≠ f(b)), then there exists a real number c in the interval (a, b) such that f(c) = v.

  The proof of the Intermediate Value Theorem (IVT) uses methods of real analysis that are too advanced to be presented here, so the proof is omitted.

  REMARK 7.7.1. Loosely speaking, the IVT states that, if the domain of a continuous real-valued function is a closed interval I, then the range of the function always contains the closed interval determined by the values of the function at the end points of the interval I. For example, if the domain of a continuous real-valued function is the closed interval [2, 3], then its range contains the interval [f(2), f(3)] or [f(3), f(2)], depending on whether f(2) < f(3) or f(3) < f(2). (In the case that f(2) = f

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  eBook - ePub

  Understanding Real Analysis

  • Paul Zorn(Author)
  • 2017(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  4.3 The Mean Value Theorem

  The mean value theorem (MVT) is a marquee result of real analysis. As the Intermediate Value Theorem (IVT) and the extreme value theorem (EVT) do for continuous functions, the mean value theorem addresses a natural and fundamental property of differentiable functions.

  Figure 4.5 The mean value theorem illustrated.

  Theorem 4.9 (The mean value theorem). Suppose

  f : [ a , b ] → R

  is continuous on

  [ a , b ]

  and differentiable on

  ( a , b )

  . Then there exists

  c ∈ ( a , b )

  such that

  f ′

  ( c )

  =

  f ( b ) - f ( a )

  b - a

  .

  Like the IVT and the EVT, the MVT says that a certain function–f ′ , this time—assumes a certain “mean” (or average”) value. Unexciting as this might seem, it turns out to be a big deal, with surprisingly broad and deep implications. We explore several before proving the theorem.

  The graphical view. The MVT equation says something about slopes on the graph of f :At some point c between a and b , the tangent line is parallel to the secant line joining

  ( a , f ( a ) )

  and

  ( b , f ( b ) )

  . Figure 4.5 illustrates this. The (linear) secant line function, labeled L (x ) in the picture, will play a role in the proof.

  The horizontal case: Rolle’s theorem. If we add to the other hypotheses the requirement that f (a ) = f (b ), then the conclusion takes a slightly simpler form:

  Theorem 4.10 (Rolle’s theorem). Suppose

  f : [ a , b ] → R

  is continuous on

  [ a , b ]

  and differentiable on

  ( a , b )

  , and that f (a ) = f (b ) . Then f ’ (c ) = 0 for some

  c ∈ ( a , b )

  .

  In graphical terms, the new hypothesis says that the graph has the same height at x  = a and at x  = b ; the conclusion says that the graph has a horizontal tangent somewhere in between. Figure 4.6 shows two such horizontal tangents. More than one possible c is fine with Rolle.

  Figure 4.6 Rolle’s theorem: two possible c ‐values.

  Car talk. If f (t ) is the position of a car at time t

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  eBook - PDF

  A Basic Course in Real Analysis

  • Ajit Kumar, S. Kumaresan(Authors)
  • 2014(Publication Date)
  • Chapman and Hall/CRC

    (Publisher)

  Hence it follows that lim a n = c = lim b n . Since c ∈ J and f is continuous on J , we have f ( a n ) → f ( c ) and f ( b n ) → f ( c ) . Since f ( a n ) ≤ 0 for all n , it follows that lim n f ( a n ) ≤ 0, that is, f ( c ) ≤ 0, (by the analogue of Item 2 in Exercise 2.1.28). In an analogous way, f ( c ) = lim f ( b n ) ≥ 0. We are forced to conclude that f ( c ) = 0. The proof is complete. 3.3. Intermediate Value Theorem 81 Theorem 3.3.2 (Intermediate Value Theorem – Standard Version) . Let g : [ a, b ] → R be a continuous function. Let λ be a real number between g ( a ) and g ( b ) . Then there exists c ∈ ( a, b ) such that g ( c ) = λ . Proof. Strategy: Apply the previous version to the function f ( x ) = g ( x ) -λ . Assume without loss of generality, g ( a ) < λ < g ( b ). Then f ( a ) < 0 and f ( b ) > 0. Also g is continuous on [ a, b ]. Hence, by Theorem 3.3.1, there exists a c ∈ ( a, b ) such that f ( c ) = 0. That is, g ( c ) = λ . Remark 3.3.3. We made a crucial use of the LUB property of R in the proofs of the theorems above. They are not true, for example, in Q . We shall be brief. Consider the interval [0 , 2] ∩ Q and the continuous function f ( x ) = x 2 -2. Then f (0) < 0 while f (2) > 0. We know that there exists no rational number α whose square is 2. Recall also that we have shown that Q does not enjoy the LUB property (Remark 1.3.20). Exercise Set 3.3.4. Three typical applications of the intermediate value theo-rem. (1) Let J ⊂ R be an interval. Let f : J → R be continuous and f ( x ) 6 = 0 for any x ∈ J . Show that either f > 0 on J or f < 0 on J . (2) Let f : R → R be continuous taking values in Z or in Q . Then show that f is a constant function. (3) Let f : [ a, b ] → R be a non-constant continuous function. Show that f ([ a, b ]) is uncountable. Remark 3.3.5. The Intermediate Value Theorem says that the image of an in-terval under a continuous function is again an interval. Let J ⊂ R be an interval. Let f : J → R be continuous.

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  Calculus

  Resequenced for Students in STEM

  • David Dwyer, Mark Gruenwald(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  Thus, f is continuous everywhere that it is defined, and so we may evaluate its limit by substitution. lim x→0 4e x - sin 2 x x 3 + √ x + 4 = f (0) = 4e 0 - sin 2 0 0 3 + √ 0 + 4 = 4 - 0 0 + 2 = 4 2 = 2 The Intermediate Value Theorem Suppose that it is 70 ◦ at 1 P.M and 80 ◦ at 4 P.M. Our intuition tells us that temperatures don’t skip values, and so we would expect that the temperature would hit every value between 70 ◦ and 80 ◦ at some time in between. That is, we would expect a temperature function to attain every value intermediate between 70 and 80 at some point on the interval from 1 to 4. We believe this because we know from experience (like reading a mercury thermometer) that temperature may rise or fall rapidly, but it does so continuously, without jumping over or omitting values along the way. More generally, all continuous functions have the property that they don’t “skip” values. That is, they attain all values between any pair of function values. This is formalized in the Intermediate Value Theorem, the proof of which requires material beyond the scope of this course. Theorem 2.5.5 Intermediate Value Theorem If f is continuous on the closed interval [a, b] and N is any number between f (a) and f (b) (with f (a) 6= f (b)), then there is a number c in (a, b) such that f (c) = N . This means that the graph of f crosses the horizontal line y = N at a point (c, f (c)) somewhere between (a, f (a)) and (b, f (b)), as shown in Figure 2.43. f (a) N f (b) a c b y = f (x) y = N Figure 2.43 Example 7 Applying the Intermediate Value Theorem Show that the equation cos x = x has a solution on the interval [0, π 2 ]. 98 CHAPTER 2. LIMITS Solution If we evaluate cos x and x at the endpoints of the interval, we find that cos 0 = 1 > 0, but cos π 2 = 0 < π 2 In other words, the inequality reverses: at the left endpoint cos x > x, but at the right endpoint cos x < x.

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  Advanced Calculus

  An Introduction to Linear Analysis

  • Leonard F. Richardson(Author)
  • 2011(Publication Date)
  • Wiley-Interscience

    (Publisher)

  Proof: By hypothesis if E > 0 there exists 8 > 0 such that IIPII < 8 implies Let P = { xo, ... , x n } be any partition of [a, b] such that I IPII < 8. By the Mean Value Theorem in each sub-interval [Xi -1, Xi] we are free to pick Xi in such a way that F'(Xi)6.xi = F(Xi) -F(Xi-d. Then n i= l n = L F' (Xi)6. xi i= l n i= l = F(b) -F(a). Since this implies IF(b ) -F(a) -J : f(x) dxl < E for all E > 0, we must have J: f( x) dx = F(b) -F(a). -• EXAMPLE 4.4 Let { 2' 1 F(x)= ~ Slllx2 ifO

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  The How and Why of One Variable Calculus

  • Amol Sasane(Author)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  Corollary 3.7. Let I ⊂ R be an interval and f : I → R be continuous. Then f ( I ) is an interval. shadow of graph is an interval shadow is not an interval I I 106 THE HOW AND WHY OF ONE VARIABLE CALCULUS Proof. Let y 1 , y 2 ∈ f ( I ) be such that y 1 < y 2 . Then there exist a , b ∈ I such that y 1 = f ( a ) , y 2 = f ( b ) . Let y ∈ [ y 1 , y 2 ] . By the Intermediate Value Theorem, there exists a point c , belong-ing to the compact interval with endpoints a and b , such that f ( c ) = y . So y ∈ f ( I ) . As the choice of y ∈ [ y 1 , y 2 ] was arbitrary, it follows that [ y 1 , y 2 ] ⊂ f ( I ) . Hence f ( I ) has the interval property, and consequently, f ( I ) is an interval. The above corollary can be summarised by saying that ‘continuous functions preserve connectedness’. Exercise 3.34. In each of the following cases, give an example of a continuous function f : S → R such that f ( S ) = T , or explain why such an f can’t exist. (1) S = ( 0, 1 ) , T = ( 0, 1 ] . (2) S = ( 0, 1 ) , T = { 0, 1 } . 3.4 Extreme Value Theorem Theorem 3.8 (Extreme Value Theorem) . If f : [ a , b ] → R is continuous, then (1) S : = { f ( x ) : x ∈ [ a , b ] } = : f ([ a , b ]) = range of f is bounded. (2) sup S and inf S exist. (3) sup S and inf S are attained, that is, there exist c , d ∈ [ a , b ] such that f ( c ) = sup S = max S and f ( d ) = inf S = min S. Thus, in the above conclusion, we have f ( c ) ≥ f ( x ) for all x ∈ [ a , b ] (so that c is a maximiser of f ), and we have f ( d ) ≤ f ( x ) for all x ∈ [ a , b ] (so that d is a minimiser of f ). Note that continuity of f says something locally about f at each point of its domain. How-ever, the conclusion says something globally about f . This miracle happens because [ a , b ] is ‘compact’. We will later see examples that show that maximisers/minimisers may fail to exist if either [ a , b ] is not compact or if f is not continuous. First, let us prove the Extreme Value Theorem. Proof.

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