Employ the following methods to find the maximum of the function from Prob.13.8: (a) Golden-section search (x1 = –2, xu = 1, es = 1%). (b) Quadratic interpolation (x0 = –2, x1 = –1, x2 = 1, iterations...

(a) First, the golden ratio can be used to create the interior points,
8541.1))2(1(
2
15



d
1459.08541.121 x
8541.08541.112 x
The function can be evaluated at the interior points
8514.0)8541.0()( 2  fxf
5650.0)1459.0()( 1  fxf
Because f(x1) > f(x2), the maximum is in the interval defined by x2, x1 and xu where x1 is
the optimum. The error at this point can be computed as
%41.785%100
1459.0
)2(1
)61803.01( 


a
The process can be repeated and all the iterations summarized as

i xl f(xl) x2 f(x2) x1 f(x1) xu f(xu) d xopt a
1 -2 -22 -0.8541 -0.851 -0.1459 0.565 1 -16.000 1.8541 -0.1459 785.41%
2 -0.8541 -0.851 -0.1459 0.565 0.2918 -2.197 1 -16.000 1.1459 -0.1459 485.41%
3 -0.8541 -0.851 -0.4164...