Question

9. Say we have a function (x+5) (x-6). If we start Newton-Ra what are the next two x values? x-2, 10·Given the function in # 9 a. Is there a minimum or maximum on the interval -2 to 1 b. What is the value? (analytical solution) Describe how you could use the bisection method or newton to find it numerically just describe the method in some detail. c.
answer 10 please

Answer 1

f (z) = (x + 5) times (x - 6) = x^2 - x 0 - 30 differentiating above f^n w. r. t x, we have d/dx f (x) = d/dx (x^2 - x - 30) = 2x - 1 Now put f' (x) = 0, thus 2x - 1 = 0 or x = 1/2 f (1/2) = (1/2)^2 - 1/2 - 30 = 1/4 - 12 - 30 = -1/2 - 30 = -61/2 Since f is defined on interval [- 2, 10], = -30.5 thus f (- 2) = (- 2)^2 - (- 2) - 30 = 4 + 2 - 30 = -24 f (10) = 10^2 - (10) - 30 = 100 - 40 = 60 (a) Yes, there is max^m and min^m values (b) Maximum value is - 30.5 at point (1/2) (c) since Newton's method is given by x_n + 1 = x_n - f (x_n)/f' (x_n) (*) we have to assume some initial approximation say x_0 = 'a' Then after we just have to find the value's pf above expression (*) that leads to more approximate root of the given function. { Assume we find x_1, x_2.. as we find more value, we go more lousier t = o the result

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