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Unformatted text preview: 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 9 18.01 Fall 2006 Lecture 9: Linear and Quadratic Approximations Unit 2: Applications of Differentiation Today, we’ll be using differentiation to make approximations. Linear Approximation y=f(x) y = b+a(x-x ) y x b = f(x ) ; x ,f(x ) ( ) a = f’(x ) Figure 1: Tangent as a linear approximation to a curve The tangent line approximates f ( x ). It gives a good approximation near the tangent point x . As you move away from x , however, the approximation grows less accurate. f ( x ) ≈ f ( x ) + f ( x )( x − x ) Example 1. f ( x ) = ln x, x = 1 (basepoint) 1 f (1) = ln 1 = 0; f (1) = = 1 x x =1 ln x Change the basepoint: Basepoint u = x − 1 = 0. ≈ f (1) + f (1)( x − 1) = + 1 · ( x − 1) = x − 1 x = 1 + u = ⇒ u = x − 1 ln(1 + u ) ≈ u 1 Basic list of linear approximations In this list, we always use base point x = 0 and assume that | x | << 1 . 1. sin x ≈ x (if x ≈ 0) (see part a of Fig. 2) 2. cos x ≈ 1 (if x ≈ 0) (see part b of Fig. 2) x 3. e ≈ 1 + x (if x ≈ 0) 4. ln(1 + x ) ≈ x (if x ≈ 0) 5. (1 + x ) r ≈ 1 + rx (if x ≈ 0) Proofs Proof of 1: Take f ( x ) = sin x , then f ( x ) = cos x and f (0) = f (0) = 1 ,f ( x ) ≈ f (0) + f (0)( x − 0) = 0 + 1 .x So using basepoint x = 0 ,f ( x ) = x . (The proofs of 2, 3 are similar. We already proved 4 above.) Proof of 5: f ( x ) = (1 + x ) r ; f (0) = 1 f (0) = d (1 + x ) r x =0 = r (1 + x ) r − 1 x =0 = r dx | | f ( x ) = f (0) + f (0) x = 1 + rx y = x sin(x)...
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This note was uploaded on 01/18/2012 for the course MATH 18.01 taught by Professor Brubaker during the Fall '08 term at MIT.

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lec9 - MIT OpenCourseWare http/ocw.mit.edu 18.01 Single...

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