Rolle's Theorem

8 Key excerpts on "Rolle's Theorem"

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  The Fundamentals of Mathematical Analysis

  • G. M. Fikhtengol'ts, I. N. Sneddon(Authors)
  • 2014(Publication Date)
  • Pergamon

    (Publisher)

  Rolle's Theorem. Suppose that (1) the function f(x) is defined and is continuous in the closed interval [a, b], (2) there exists a finite derivative f(x at least in the open interval (a, b)t, (3) at the ends of the interval the function takes equal values, i.e. f(a) = f(b). Under such circumstances, between a and b a point c can be found (am. We know that both these values of the function are attained, but since/(a) =f(b) they cannot both occur on the bound-aries of the interval, and at least one of them occurs at a point c between a and b. Therefore it follows from Fermat's theorem that the derivative/'(c) vanishes at this point. This completes the proof of the theorem. In the language of geometry Rolle's Theorem states the following: if the extreme ordinates of the curve y = f(x) are equal, a point t Michel Rolle (1652-1719)—a French mathematician who for a long time was opposed to the new calculus and adopted it only at the end of his life. This theorem was announced by him for polynomials. * Obviously, the continuity of the function f(x) in (a, b) follows from (2), but neither here nor later shall we attempt to split up the condition of the theorem into independent assumptions. 186 6. THEOREMS OF DIFFERENTIAL CALCULUS can be found on the curve at which the tangent is parallel to the x-axis (Fig. 39). We emphasize that the continuity of the function f(x) in the closed interval [a, b] and the existence of the derivative in the whole open interval (a, b) are essential for the validity of the theorem.

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  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)

  This example also emphasizes the requirement of continuity at the end points of the closed interval . [Now refer to Note (2) and Figure 20.3 .]

  20.2.1 Geometric Conclusion of Rolle's Theorem

  Rolle's Theorem says (geometrically) that a curve that rises and falls (without any breaks or slopeless points ) must have leveled off in the mean time.

  20.2.2 Dynamic Face to Rolle's Theorem

  When a ball is thrown up vertically at instant t = a (say), and returns at t = b , there is an instant c , between a and b at which the ball stops momentarily, that is, it has zero velocity.

  20.2.3 A Useful Interpretation of Rolle's Theorem

  Rolle's Theorem gives us the fact that if a polynomial has n distinct zeros, its derivative has at least n − 1 distinct zeros .

  Consider a polynomial equation f (x ) = 0 where f (x ) satisfies the conditions of Rolle's Theorem , and let x 1 , x 2 , x 3 , . . ., x n be the roots of the equation. Then, by Rolle's Theorem, the equation f ′(x ) = 0 has the roots c 1 , c 2 , c 3 , . . ., c

  n −1

  , one or more of which lie in between the roots of f (x ) = 0, that is, x 1 < c 1 < x 2 < c 2 < x 3 < x

  n −1

  < c

  n −1

  < x n .

  [We studied maxima and minima (exterma) of a function in earlier Chapter 19b and found that c 1 , c 2 , c 3 , . . ., c n

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  Mathematica for Physicists and Engineers

  • K. B. Vijaya Kumar, Antony P. Monteiro(Authors)
  • 2023(Publication Date)
  • Wiley-VCH

    (Publisher)

  Then there exists some c in (a, b), such that f ′ (c) = 0. Example 10.37 Verify Rolle’s theorem for f (x) = x 2 + 2x − 8, x  [−4,2]. Solution In[105] := f[x − ] = x 2 + 2 x − 8; In[106] := f[−4] Out[106] = 0 In[107] := f[2] Out[107] = 0 Hence, the value of f (x) at x = − 4 and x = 2 coincide. Hence, according to Rolle’s theorem, there is a point c ϵ (−4,2), where f ′ (c)=0. In[108] := Solve[f ′ [x] = = 0, x] Out[108] = {{x → −1}} Hence c = −1 In[109] := Plot[{f[x], f[−1]}, {x, − 4, 2}] Out[109] = From the graph, the slope of the curve becomes zero at x = − 1. This is precisely the meaning of Rolle’s theorem, the slope of the tangent at any point of the graph f (x) is nothing but the derivative of f (x) at that point. 10.6 Mean Value Theorem If f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). Then there exists some c in (a, b), such that f ′ (c) = [ f (b)−f (a) b−a ] . Example 10.38 Verify mean value theorem if f (x) = x 3 − 5 x 2 − 3x, in the interval [a, b], where a = 1 and b = 3. Find all c  [1, 3] for which f ′ (c) = 0. 10.6 Mean Value Theorem 199 –4 –3 –2 –1 1 2 –8 –6 –4 –2 Figure 10.1 Plot of f (x) and f (−1) where f (x) = x 2 + 2x − 8. Solution In[110] := f[x − ] = x 3 − 5 x 2 − 3 x; In[111] := a = 1; In[112] := b = 3; In[113] := f ′ [x] Out[113] = −3 − 10 x + 3 x 2 f ′ (x) exists in (1, 3), hence derivable. Conditions of the mean value theorem are satisfied. Hence, there are at least some c  (1, 3) such that f ′ (c) = [ f (3)−f (1) 3−1 ] . Let, In[114] := m = f[b]−f[a] b−a Out[114] = −10 In[115] := Plot[f ′ [x] − m, {x, 1, 3}] Out[115] = To determine the roots of f ′ (x) − m, we solve the equation. In[116] := Solve[f ′ [x] − m == 0, x] Out[116] = {{x → 1}, {x → 7 3 }} The c = 7 3  (1,3), hence the mean value theorem is satisfied. In[117] := Plot[{f ′ [x], f [   ] , f[], f[]}, {x, , }] Out[117] = Below is the geometric interpretation of the mean value theorem.

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  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  In the right-hand picture, f ( a ) = f ( b ) and the function is differentiable on ( a, b ), but it isn’t continuous on all of [ a, b ]: the point x = a spoils everything. Once again, no Rolle’s Theorem allowed. Here’s an example of an application of Rolle’s Theorem. Suppose that you have a function f satisfying f 0 ( x ) > 0 for all x . In Section 10.1.1 in the previous chapter, we claimed that f must satisfy the horizontal line test. Let’s prove this using Rolle’s Theorem, arguing by contradiction. Start off by supposing that f does not satisfy the horizontal line test. Then there’s some horizontal line, say y = L , which intersects the graph of y = f ( x ) twice (or more). Suppose that two of these intersection points have x -coordinates a and b . So we know that f ( a ) = L and f ( b ) = L . In particular, f ( a ) = f ( b ), and we can use Rolle’s Theorem (we already know that f is differentiable everywhere, so it must be continuous everywhere as well). The theorem says that there is some c between a and b such that f 0 ( c ) = 0. This is impossible because f 0 ( x ) is always supposed to be positive! So the horizontal line test does not fail. Now, let’s look at an even harder example. Suppose now that the second derivative of f exists everywhere and that f 00 ( x ) > 0 for all real x . The problem is to show that f has at most two x -intercepts. Before we tackle the problem itself, let’s just think about what it means for a second or two. Can you think of a function f with f 00 ( x ) > 0 for all x that has no x -intercepts? How about one x -intercept? Two x -intercepts? If you can do all these, then try to find one with three x -intercepts. Don’t spend too long on this one, though, because it’s impossible! Indeed, our problem is to show that you can’t have more than two x -intercepts.

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  Calculus of a Single Variable: Early Transcendental Functions, International Metric Edition

  • Ron Larson, Bruce Edwards(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Rolle’s Theorem The function f (x) = braceleft.alt1 0, 1 - x, x = 0 0 < x ≤ 1 is differentiable on (0, 1) and satisfies f (0) = f (1). However, its derivative is never zero on (0, 1). Does this contradict Rolle’s Theorem? Explain. 68. Mean Value Theorem Can you find a function f such that f (-2) = -2, f (2) = 6, and f uni2032 (x) < 1 for all x? Why or why not? 220 Chapter 4 Applications of Differentiation 70. Temperature When an object is removed from a furnace and placed in an environment with a constant temperature of 32°C, its core temperature is 820°C. Five hours later, the core temperature is 200°C. Explain why there must exist a time in the interval (0, 5) when the temperature is decreasing at a rate of 124°C per hour. 71. Velocity Two bicyclists begin a race at 8:00 a.m. They both finish the race 2 hours and 15 minutes later. Prove that at some time during the race, the bicyclists are traveling at the same velocity. 72. Acceleration At 9:13 a.m., a sports car is traveling 55 kilometers per hour. Two minutes later, the car is traveling 135 kilometers per hour. Prove that at some time during this two-minute interval, the car’s acceleration is exactly 2400 kilometers per hour squared. 73. Think About It Sketch the graph of an arbitrary function f that satisfies the given condition but does not satisfy the conditions of the Mean Value Theorem on the interval [-5, 5]. (a) f is continuous. (b) f is not continuous. 74. HOW DO YOU SEE IT? The figure shows two parts of the graph of a continuous differentiable function f on [-10, 4]. The derivative f uni2032 is also continuous. x - 8 - 4 4 8 4 - 4 - 8 y (a) Explain why f must have at least one zero in [-10, 4]. (b) Explain why f uni2032 must also have at least one zero in the interval [-10, 4]. What are these zeros called? (c) Make a possible sketch of the function, where f uni2032 has one zero on the interval [-10, 4].

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  Real Analysis and Applications

  Including Fourier Series and the Calculus of Variations

  • Frank Morgan(Author)
  • 2005(Publication Date)
  • American Mathematical Society

    (Publisher)

  Part III Calculus Part III: Calculus. It is satisfying to see how naturally the theory of calculus follows from the basic concepts of real analysis, especially when you remember that after Newton and Leibniz invented the calculus in the late 1600s, it took mathematicians two hundred years to get the theory straight. This part focuses on R 1 rather than R n , although Chapters 16 and 17 hold for real-valued functions on R n as well. Chapter 13 The Derivative and the Mean Value Theorem Differentiation and integration are the two distinguishing processes of calcu-lus. The first major theorem about the derivative, the Mean Value Theorem, follows easily from the compactness of the interval [ a, b ], via Proposition 13.2 and Rolle’s Theorem 13.3. We begin by recalling the definition of the deriv-ative of a function on an interval in R . 13.1. Definition. Let f be a real-valued function on an open interval ( a, b ). Define the derivative f = df /dx by f ( x ) = lim t → x f ( t ) − f ( x ) t − x = lim ∆ x → 0 ∆ f ∆ x . If this limit exists, we say that f is differentiable at x . One important interpretation of the derivative is the slope of the graph. More generally, the derivative gives a rate of change. We assume the easy and familiar properties of the derivative and now state and prove the important theorems. We mention that if f is differen-tiable at x then f is continuous at x . If f has a continuous derivative, we say that f is continuously differentiable or C 1 . If f has k continuous derivatives, we say that f is C k . If f has derivatives of all orders, we say that f is C ∞ or smooth. (Sometimes smooth is used for C 1 . It is a somewhat vague term that means one does not want to worry about differentiability and wants to assume whatever is needed.) We say that f is piecewise C k or smooth if the domain is the union of finitely many closed intervals on which f is C k or smooth. (For this 61

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  Advanced Calculus

  An Introduction to Linear Analysis

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

    (Publisher)

  (See Fig. 4.2.) Observe that this is a generalization of Rolle's Theorem. Proof: Observe that the straight line that is the chord joining the endpoints of the graph of f is described by the equation f(b) -f(a) y = f(a) + (x -a). b -a Now, for all x E [a, b], let f(b) -f(a) h(x) = f(x)-y = f(x)-f(a)-b -a (x-a) , THE MEAN VALUE THEOREM 107 [h] Figure 4.2 f (x) = ~, chord with parallel tangent. the difference in height between the graph of f and the graph of the chord. Observe that hE C[a, b] and h(a) = 0 = h(b). Also, hi exists at least on (a, b). Thus Rolle 's Theorem implies there exists p, E (a, b) such that hi (p,) = O. This implies that • Corollary 4.2.1 If fl == 0 on an interval I, then f is a constant function on I. Proof: Fix any a E I and let x E I be arbitrary. Then the Mean Value Theorem implies there exists p, E I such that f(x) -f(a) = f'(p,) (x -a) = O(x -a) = 0 so that f (x) == f (a), a constant function. • Corollary 4.2.2 If l' == gl on an interval I, then there exists c E lR such that f(x) == g(x) + c. Proof: Let h(x) = f(x) -g(x) and observe that hi == 0 on I. Now use Corollary 4.2.1. • Corollary 4.2.3 Suppose fEe [a, b] is differentiable at least on (a, b). Then 108 THE DERIVATIVE i. If f'(x) 2': Of or all x E (a, b), then f is increasing (denoted by f /) on [a, b]. ll . If f' (x) > 0 for all x E (a, b), then f is strictly increasing (denoted by f r) on [a , b]. iii. Iff' (x) ::; 0 for all x E (a, b), then f is decreasing (denoted by f -.,,) on [a, b ]. iv. If f' (x) < 0 for all x E (a, b), then f is strictly decreasing (denoted by f 1) on [a , b ]. Proof: i. Suppose that f'(x) 2': 0 for all x E (a, b). Then if Xl < X2 are points of [a, b], the Mean Value Theorem implies there exists fJ E (a, b) such that so that f(X2) 2': f(XI), and f is increasing on [a, b]. For the remaining cases, see Exercises 4.10-4.12. • EXAMPLE 4.2 If x> 0, then s in x < x. Proof: Let f( x) = x -sinx. Then f(O) = 0 and f'( x) = 1 -cosx > 0 on (0, 27f).

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  Theoretical Introduction to Mathematical Theorems, A

  • (Author)
  • 2014(Publication Date)
  • Learning Press

    (Publisher)

  In higher dimensions Lebesgue's differentiation theorem generalizes the Fundamental theorem of calculus by stating that for almost every x , the average value of a function ƒ over a ball of radius r centered at x will tend to ƒ ( x ) as r tends to 0. Part II of the theorem is true for any Lebesgue integrable function ƒ which has an antiderivative F (not all integrable functions do, though). In other words, if a real function F on [ a , b ] admits a derivative ƒ ( x ) at every point x of [ a , b ] and if this derivative ƒ is Lebesgue integrable on [ a , b ], then Rudin (1987, th. 7.21) This result may fail for continuous functions F that admit a derivative ƒ ( x ) at almost every point x , as the example of the Cantor function shows. But the result remains true if F is absolutely continuous: in that case, F admits a derivative ƒ ( x ) at almost every point x and, as in the formula above, F ( b ) − F ( a ) is equal to the integral of ƒ on [ a , b ]. ________________________ WORLD TECHNOLOGIES ________________________ The conditions of this theorem may again be relaxed by considering the integrals involved as Henstock-Kurzweil integrals. Specifically, if a continuous function F ( x ) admits a derivative ƒ ( x ) at all but countably many points, then ƒ ( x ) is Henstock-Kurzweil integrable and F ( b ) − F ( a ) is equal to the integral of ƒ on [ a , b ]. The difference here is that the integrability of ƒ does not need to be assumed. (Bartle 2001, Thm. 4.7) The version of Taylor's theorem which expresses the error term as an integral can be seen as a generalization of the Fundamental Theorem. There is a version of the theorem for complex functions: suppose U is an open set in C and ƒ : U → C is a function which has a holomorphic antiderivative F on U . Then for every curve γ : [ a , b ] → U , the curve integral can be computed as The fundamental theorem can be generalized to curve and surface integrals in higher dimensions and on manifolds.

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