12 + 2 = 1 -S = 1.2 =} X3 = 1.2 -3(1.2)2 + 2 ~ 1.1797. 7. f(x) = x4-20 , ) 3 f(xn) x; -20 =} j (x = 4x , so Xn+l = Xn-f'(xn) = Xn-. -· Now x1 = 2 =?--24 -20 --(2.125)4 -20 ,..._, X2 -2-4(2)3 -2.125 =} X3 -2.125-4(2.125)3 ,..._, 2.1148. 9. f(x) = x3 + x + 3 =?-f' (x) = 3x2 + 1, so X~ +xn +3 Xn+l = Xn -3 2 1 . Now X1 = -1 =?-Xn + (-1)3+(-1)+3 -1-1+3 1 X2 = -1-= -1-= -1--= -1.25. 3(-1)2+1 3+1 4 Newton's method follows the tangent line at ( -1, 1) down to its intersection with the x-axis at ( -1.25, 0), giving the second approximation x2 = - 1.25. -4 11. To approximate x = .V3Q (so that x3 = 30), we can take f ( x) = x3 -30. So f' ( x) = 3x2, and thus, Xn+l = Xn-x~3 -2 30. Since -V27 = 3 and 27 is close to 30, we'll use x1 = 3. We need to find approximations Xn until they agree to eight decimal places. x1 = 3 =?-x2 ~ 3.11111111, X3 ~ 3.10723734, X4 ~ 3.10723251 ~ xs. So .V3Q ~ 3.10723251, to eight decimal places. Here is a quick and easy method for finding the iterations for Newton's method on a programmable calculator. (The screens shown are from the TI -83 Plus, but the method is similar on other calculators.) Assign f ( x) = x3 -30 toY 1. and f' (x) = 3x2 toY 2· Now store x1 = 3 in X and then enter X-Y 1/Y 2 --+ X to get x2 = 3.1. By successively pressing the ENTER key, you get the approximations X3, X4, ....tl~ti tl~t2 tl~t~ -...Y1EIX~-30 -...Y2EI3X2 ,y~=· 'Y~t= -...Ys:= -...Y&= -...Y7= 3~X 3 x-Y1/Y2~X 3.111111111 3.107237339 3.107232506 3.107232506 I In Derive, load the utility file SOLVE. Enter NEWTON ( x" 3-3 0 , x, 3) and then APPROXIMATE to get [3, 3.11111111, 3.10723733, 3.10723250, 3.10723250]. You can request a specific iteration by adding a fourth argument. For example, NEWTON (x"3 -30, x, 3, 2) gives [3, 3.11111111, 3.10723733]. In Maple, make the assignments f := x--+ x A3-30;, g := x--+ x-f(x)/ D(f)(x);, and x := 3.;. Repeatedly execute the command x := g(x); to generate successive approximations. In Mathematica, make the assignments f[x_] := x A3 ~ 30, g[x_] := x-f[x]/ f'[x], and x = 3. Repeatedly execute the command x = g[x] to generate successive approximations.

194 D CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 13. f(x) = 2x3-6x2 + 3x + 1 ::::} j'(x) = 6x2-12x + 3 2x~ -6x; + 3xn + 1 . We ::::} Xn+I = Xn -6x?t -12Xn + 3 need to find approximations until they agree to six decimal places. XI = 2.5 ::::} x2 ~ 2.285714, X3 ~ 2.228824, X4 ~ 2.224765, xs ~ 2.224745 ~ X6. So the root is 2.224745, to six decimal places. 15. sinx = x2, so f(x) = sinx-x2 ::::} j'(x) = cosx-2x ::::} 17. 19. . 2 Sin Xn -Xn F th fi h . . f Xn+I = Xn -2 . rom e gure, t e positive root o COSXn-Xn sin x = x2 is near 1. XI = 1 ::::} x2 ~ 0.891396, X3 ~ 0.876985, X4 ~ 0.876726 ~ xs. So the positive root is 0.876726, to six decimal places. X 3 From the graph, we see that there appear to be points of intersection near x = -0.7 and x = 1.2. Solving x4 = 1 + x is the same as solving f(x) = x4-x -1 = 0. f(x) = x4-x-1 ::::} f'(x) = 4x3-1, I ;ole > I < I 12 X~-Xn -1 SO Xn+I = Xn-4x~-1 -1 XI= -0.7 X2 ~ -0.725253 X3 ~ -0.724493 X4 ~ '-0.724492 ~ X5 XI= 1.2 X2 ~ 1.221380 X3 ~ 1.220745 X4 ~ 1.220744 ~ X5 To six decimal places, the roots of the equation are -0.724492 and 1.220744. 3 -31 X '-, 13 -3 From the graph, there appears to be a point of intersection near x = 0.5. Solving tan-I x = 1-xis the same as solving f(x) = tan- I x + x-1 = 0. f(x) = tan-I x + x-1 ::::} 1 1 tan -I Xn + Xn -1 f (x) = -1--2 + 1, SO Xn+I = Xn-. . _. . +x XI = 0.5 ::::} X2 ~ 0.520196, X3 ~ 0.520269 ~ X4. So to six decimal places, the root of the equation is 0.520269. 21. From the graph, there appears to be a point of intersection near x = 0.6. 2 Solving cosx = Jx is the same as solving f(x) = cosx-Jx = 0.

#### You've reached the end of your free preview.

Want to read all 87 pages?

- Fall '19