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MIT 18 01 - Mean Value Theorem and Inequalities

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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 14 18.01 Fall 2006 Lecture 14: Mean Value Theorem and Inequalities Mean-Value Theorem The Mean-Value Theorem (MVT) is the underpinning of calculus. It says: If f is differentiable on a < x < b, and continuous on a ≤ x ≤ b, then f(b) − f(a) = f�(c) (for some c, a < c < b)b − a f(b) − f(a)Here, is the slope of a secant line, while f�(c) is the slope of a tangent line. b − a secant lineslope f’(c)abcFigure 1: Illustration of the Mean Value Theorem. Geometric Proof: Take (dotted) lines parallel to the secant line, as in Fig. 1 and shift them up from below the graph until one of them first touches the graph. Alternatively, one may have to start with a dotted line above the graph and move it down until it touches. If the function isn’t differentiable, this approach goes wrong. For instance, it breaks down for the function f(x) = |x|. The dotted line always touches the graph first at x = 0, no matter what its slope is, and f�(0) is undefined (see Fig. 2). 1Lecture 14 18.01 Fall 2006 Figure 2: Graph of y = |x|, with secant line. (MVT go es wrong.) Interpretation of the Mean Value Theorem You travel from Boston to Chicago (which we’ll assume is a 1,000 mile trip) in exactly 3 hours. At 1000some time in between the two cities, you must have been going at exactly mph.3 f(t) = position, measured as the distance from Boston. f(3) = 1000, f(0) = 0, a = 0, and b = 3. 1000 = f(b) − f (a)= f�(c)3 3 where f�(c) is your speed at some time, c. Versions of the Mean Value Theorem There is a second way of writing the MVT: f(b) − f (a) = f�(c)(b − a) f(b) = f (a) + f�(c)(b − a) (for some c, a < c < b) There is also a third way of writing the MVT: change the name of b to x. f(x) = f (a) + f�(c)(x − a) for some c, a < c < x The theorem does not say what c is. It depends on f , a, and x. This version of the MVT should be compared with linear approximation (see Fig. 3). f(x) ≈ f(a) + f�(a)(x − a) x near a 2Lecture 14 18.01 Fall 2006 The tangent line in the linear approximation has a definite slope f�(a). by contrast formula is an exact formula. It conceals its lack of specificity in the slope f�(c), which could be the slope of f at any point between a and x. (a,f(a))(x,f(x))y=f(a) + f’(a)(x-a)errorFigure 3: MVT vs. Linear Approximation. Uses of the Mean Value Theorem. Key conclusions: (The conclusions from the MVT are theoretical) 1. If f�(x) > 0, then f is increasing. 2. If f�(x) < 0, then f is decreasing. 3. If f�(x) = 0 all x, then f is constant. Definition of increasing/decreasing: Increasing means a < b f(a) < f(b). Decreasing means a < b = f(a) < f(b).⇒ ⇒ Proofs: Proof of 1: a < b f(b) = f (a) + f�(c)(b − a) Because f�(c) and (b − a) are both positive, f(b) = f (a) + f�(c)(b − a) > f (a) (The proof of 2 is omitted because it is similar to the proof of 1) Proof of 3: f(b) = f (a) + f�(c)(b − a) = f (a) + 0(b − a) = f(a) Conclusions 1,2, and 3 seem obvious, but let me persuade you that they are not. Think back to the definition of the derivative. It involves infinitesimals. It’s not a sure thing that these infinitesimals have anything to do with the non-infinitesimal behavior of the function. 3Lecture 14 18.01 Fall 2006 Inequalities The fundamental property f� > 0 = f is increasing can be used to deduce many other inequali-⇒ties. xExample. ex1. e > 0 x2. e > 1 for x > 0 x3. e > 1 + x xProofs. We will take property 1 (e > 0) for granted. Proofs of the other two properties follow: Proof of 2: Define f1(x) = ex −1. Then, f1(0) = e0 −1 = 0, and f�(x) = ex > 0. (This last assertion 1is from step 1). Hence, f1(x) is increasing, so f(x) > f(0) for x > 0. That is: e x > 1 for x > 0 . xProof of 3: Let f2(x) = e − (1 + x). f�(x) = e x − 1 = f1(x) > 0 (if x > 0).2Hence, f2(x) > 0 for x > 0. In other words, e x > 1 + x 2 2x xSimilarly, e x > 1 + x + 2 (proved using f3(x) = e x − (1 + x + 2 )). One can keep on going: 2 3x xe x > 1 + x + + for x > 0. Eventually, it turns out that 2 3! 2 3x xe x = 1 + x + + + (an infinite sum) 2 3! · · · We will be discussing this when we get to Taylor series near the end of the course.


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MIT 18 01 - Mean Value Theorem and Inequalities

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