The purpose of this problem is to learn that if you choose the initial point \"somewhat far\" from the optimal point, then the gradient methods may diverge. Consider the one-dimensional function _2 3 12 f[I]3|I| +2I' Since f(2:) Z 0 and f[0) = 0, we immediately have 3* = 0. Pick m; = 311., Where 7 is a positive scalar. We will use in the steepest descent method. Part (a) Plot the mction in MATLAB for I ranging from -5 to 5. Also compute the second derivative of the mction V2f for 1: E R\\ {0} and show that it is unbmmded? Is the function f convex? Part (13) Pick 7 = 1 and compute the expression for Ik+1 in terms of Ik. Show (without using MATLAB) that if 1 = 1 and Iinl 2 11 then lszkl 2 k + 1 for all k E N. Thus, the steepest descent method diverges if 'y = 1 and lInl 2 1. Hint: Use the principle of mathematical induction. Part (c) Use trial and error to identify a pair (1', In) for which the steepest descent method converges. You can use MATLAB to identify a pail-1. Show the rst six points In, 31, . . . , $5 on the plot using the MATLAB command terth,y, 'x_1' , 'Horizontalnlignment ' , 'left') ; or text (I ,y , 'x_2' , 'Horizontalnlignmant ' , 'right ') ; The above command puts the text :51 at the position (I, 3;) on the MATLAB. The steepest descent does not converge in this problem intense the second derivative of the function is unbounded! 1I expect everyone in the class will have a dill'erent answer from that of others for this