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Calc_Chap05

# Calc_Chap05 - MATH1131 Mathematics 1A Calculus Chapter 05...

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MATH1131 – Mathematics 1A Calculus Chapter 05 The Mean Value Theorem and Its Applications Dr. Thanh Tran School of Mathematics and Statistics The University of New South Wales Sydney, Australia 1 1 The Mean Value Theorem 2 Applications of the MVT 3 Sign of the Derivative 4 Critical, Global Max. and Min. Points 5 Counting Zeros of Functions 6 Antiderivatives 7 L’ Hôpital’s Rule 2 We are interested in questions like: If f exists on Dom ( f ) , a subset of R , can we relate values of f to values of f ? can we compare values of f to values of another differentiable function g ? can we bound errors associated with previous approximations? can we get efficient ways of dealing with max and min of functions? How to fix some indeterminate limit forms when differentiable functions are involved? All above issues, and more, are dealt with via . . . The Mean Value Theorem 3 Theorem 1 (Mean Value Theorem) Suppose that f is continuous on [ a , b ] , differentiable on ( a , b ) . Then there exists c ( a , b ) such that f ( c ) = f ( b ) f ( a ) b a . In pictures . . . The tangent to the graph of f at c is parallel to the line joining ( a , f ( a ) ) to ( b , f ( b ) ) . x b c a f(x) The Mean Value Theorem 4

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Example 2 Suppose that f : [ 1 , 4 ] R is given by f ( x ) = 3 x 4 x . Find a number c ( 1 , 4 ) that satisfies the conclusion of the MVT for f on [ 1 , 4 ] . Since f is continuous on [ 1 , 4 ] and differentiable on ( 1 , 4 ) , the MVT says there is a c ( 1 , 4 ) such that Since f ( x ) = 3 2 x 4 we have Thus c = 9 / 4. Usually we are not interested in finding the c explicitly, just knowing it exists is enough! The Mean Value Theorem 5 Using the MVT to prove inequalities Example 3 Use the mean value theorem to show that x + 4 < 2 + x 4 for x > 0. The inequality is equivalent to Let f : [ 0 , x ] R be defined by Then, since f is continuous on [ 0 , x ] and differentiable on ( 0 , x ) , by the MVT there exists c ( 0 , x ) such that Applications of the MVT 6 Since we deduce Hence x + 4 2 x 4 < 0 (noting x > 0). Applications of the MVT 7 Error bounds Example 4 Use the MVT to find an upper bound for the error involved if we approximate cos 15 π 21 by 1 / 2. Let f ( x ) = cos x . Then cos 15 π 21 = f parenleftbigg 15 π 21 parenrightbigg f parenleftbigg 2 π 3 parenrightbigg = cos 2 π 3 = cos π 3 = 1 2 . Since f is continuous on [ 2 π/ 3 , 15 π/ 21 ] and differentiable on ( 2 π/ 3 , 15 π/ 21 ) , the MVT gives Applications of the MVT 8
So an upper bound of the approximation of cos ( 15 π/ 21 ) by 1 / 2 is 3 π/ 42. MAPLE gives > evalf(abs(cos(15 * Pi * (1/21))-cos(2 * Pi * (1/3)))); 0.1234898018 > evalf((1/42) * sqrt(3) * Pi); 0.1295570975 Applications of the MVT 9 The sign of the derivative of a function tells us whether the function increases or decreases. But first we need some definitions.

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