**Problem 8:**Utilize the following methods to determine the maximum of the function:\[ f(x) = 4x - 1.8x^2 + 1.2x^3 - 0.3x^4 \]**(a) Golden-section search** - Initial lower bound (\(x_l\)): -2 - Initial upper bound (\(x_u\)): 4 - Stopping criterion (\(\varepsilon_s\)): 1%**(b) Parabolic interpolation**- Initial points: - \(x_0 = 1.75\) - \(x_1 = 2\) - \(x_2 = 2.5\) - Number of iterations: 4 - New points are selected sequentially, similar to the secant method.**(c) Newton’s method** - Initial guess (\(x_0\)): 3 - Stopping criterion (\(\varepsilon_s\)): 1% This problem involves applying numerical optimization techniques to find the maximum value of a given polynomial function. Each method provides a different approach to approximating the maximum within a specified tolerance level.

“Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.” Given fx=4x-1.8x2+1.2x3-0.3x4. Let gx=-fx. To obtain the maximum of f(x), we need the minimum of f(x) which can be obtained by Golden search method. gx=-4x+1.8x2-1.2x3+0.3x4, xl=-2, xu=4, εs=1%. Formula: d=GRxu-xlx1=xl+dx2=xu-dIf fx1<fx2 then xl=x2 and xu does not change.If fx1>fx2 then xu=x1 and xl does not change,where GR=5-12.Absolute Error: εa=2-GRxu-xlx*×100, where x*=x1+x22.Iteration 1: xl=-2xu=4d=GRxu-xl=5-124--2=3.708203932x1=xl+d=-2+3.708203932=1.708203932x2=xu-d=4-3.708203932=0.2917960675fx1=-5.007504445fx2=-1.041562423 The error is εa=2-GRxu-xlx*×100=2-5-124--21.708203932+0.29179606752×100=829.18% Iteration 2: As fx1<fx2 we have xl=x2=0.2917960675xu=4d=GRxu-xl=2.291796068x1=xl+d=2.583592135x2=xu-d=1.708203932fx1=-5.647393819fx2=-5.007504445 The error is εa=2-GRxu-xlx*×100=2-5-124-0.29179606752.583592135+1.7082039322×100=238.81% And so on we proceed.We have used spreadsheets to calculate further values, Formula: Values: We observe that on the 14th iteration εa<εs. Thus, the minimum of g(x) is the average of x1 and x2 which is x*=2.329516612. Therefore, the maximum of f(x) is x*=2.329516612.