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 UAE_Math_Grade_12_Vol_1_SE_718383_ch3
                                  GO01962-Smith-v1.cls
                                                    July 4, 2016
            4 - 1                     numerical evidence to approximate solutions. The graphing and analysis of compli-

                                                                  ximations
                                          Linear Approximations
                                          Linear Appro
                                      cated functions and the solution of equations involving these functions are the em-
                                      phases of this chapter.
                                                                 ’
                                          and Newton
                                          and Newton’s Method
                                                                  s Method
                       LINEAR APPROXIMATIONS AND NEWTON’S METHOD
                4.1
                                      There are two distinctly different tasks for which you use a scientific calculator. First,
                                      while we all know how to multiply 1024 by 1673, a calculator will give us an answer
                                      more quickly. Alternatively, we don’t know how to calculate sin(1.2345678) without a
                                      calculator, since there is no formula for sin x involving only the arithmetic operations.
                                      Your calculator computes sin(1.2345678) ≈ 0.9440056953 using a built-in program that
                                      generates approximate values of the sine and other transcendental functions.
                                         In this section, we develop a simple approximation method. Although somewhat
                                      crude, it points the way toward more sophisticated approximation techniques to follow
                                      later in the text.
                                     Linear Approximations

                                      Suppose we wanted to find an approximation for f(x ), where f(x ) is unknown, but
                                                                                             1
                                                                                  1
                                      where f(x ) is known for some x “close” to x . For instance, the value of cos(1) is
                                                                  0
                                                                             1
                                              0
                                      unknown, but we do know that cos(    ∕3) =  1 2  (exactly) and     ∕3 ≈ 1.047 is “close” to 1.
                                      While we could use  1 2  as an approximation to cos(1), we can do better.
                                         Referring to Figure 4.1, notice that if x is “close” to x and we follow the tangent
                                                                                      0
                                                                         1
                                      line at x = x to the point corresponding to x = x , then the y-coordinate of that point
                                                                              1
                                                0
                                      (y ) should be “close” to the y-coordinate of the point on the curve y = f(x) [i.e., f(x )].
                                                                                                           1
                                       1
                                                          y
                                                                         y = f(x)
                                                      f(x )
                                                        1
                                                                             y = f(x ) + f �(x )(x - x )
                                                                                             0
                                                                                  0
                                                                                       0
                                                        y 1
                                                      f(x )
                                                        0
                                                                                x
                                                                         x
                                                                   x 0    1
                                                                     FIGURE 4.1
                                                              Linear approximation of f(x )
                                                                                   1

                                                                                         ′
                                         Since the slope of the tangent line to y = f(x) at x = x is f (x ), the equation of the
                                                                                     0
                                                                                           0
                                      tangent line to y = f(x) at x = x is found from
                                                                0
                                                                            y − f(x )
                                                                      ′
                                                               m tan  = f (x ) =  x − x  0
                                                                        0
                                                                                                                    Copyright © McGraw-Hill Education
                                      or                                         0
                                                                          ′
                                                               y = f(x ) + f (x )(x − x ).                (1.1)
                                                                     0
                                                                            0
                                                                                  0
                                      We give the linear function defined by equation (1.1) a name, as follows.


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        228 | Lesson 4-1 | Linear Approximations and Newton’s Method
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