Let f(r) = e T. If we use CDF of order O(h2) with h = 0.01 to approximate f'(0), then we get %3D Select one: e0.000001 e-0.000001 O a. f'(0) z 0.02 e0.000001 0.000001 O b. f'(0) 0.0001 0.000001 0.000001 O c. f'(0) 2 0.02 21e0.000001 0.02 0.000001 O d. f'(0): O e. f'(0) = 0 0.000001 210.000001 O f. f'(0) = 0.0001 Consider the data 1, 10), (3, 6), (5,8). Then, Newton's polynomial that interpolates this data is Select one: O a. p(r) = 10 2(x 3) + 0.75(x - 3)(1 5) O b. p(x) = 10 - 2x + 0.75r? %3D O c. p(x) = 10 - 2(x- 1) + 0.75(x 1)(x-3) O d. p(r) = 10(x- 1) - 2(r 1)(x- 3) + 0.75(x 1)(x - 3)(x - 5) O e. p(r) = 10- 2(z- 10) + 0.75(z 10)(x-6)