. Using the second derivative test, classify each critical point as a local minimum, local maximum, or saddle point for the following functions. (a) f(:c) = 3:1:4 4:53 12:1:2 + 5 (b) 907; y) = 3331(1 m y) . Consider the function 8 f(a:) = 3:4 $3 6mg +6 Use Python to calculate 3:100, and its associated yvalue, that is generated by gradient descent with the following starting value 3:0 and learning rate 77: (a) 330 = 5, n = 0.01 (b) 330 = 1,:r7 = 0.01 (c) 3:0 2 1, n = 0.1 (d) 3:0 2 10, 17 = 0.0001 Note: Python may throw an UverflowError. If that happens, write it down and try to explain why it is happening (in reference to the choice of 3:0 and 77). To check that your program is working properly, if 3:0 2 4 and 17 = 0.001, then 3:100 2 3.0044144797293426 and 3/100 2 38.99953149354378. . Consider the function f(:c,y) = 5132+2y2 6:c+4y+18 (a) Sketch the graph of level curves of f(.'17,y) on the domain 10 S 3: S 10, 10 S y S 10. (b) Calculate the gradient V f . (c) Using the origin as a starting point and a learning rate of 77 = 0.1, calculate the rst ten points generated by gradient descent. Plot those ten points on your graph from part (a). [It is suggested to use Python for this part] (d) What point in the plane does it seem the ve points from part (c) are converging to (if any)