he equation f (x) = x2 − 2ex = 0 has a solution in the interval [-1,1]. With p0 = −1 and p1 = 1 calculate p2 using the Bisection With p2 from part (a) calculate p3 using Newton’s

Divide the interval of y=-1+1/2cot(2x-3pi) into 4 equal parts for both periods and calculate the correct values for the 5 essential points of Y.

Use the false position method to find the approximation to the root of the equation f(x) = x3 - 2x - 1 that is located interval [0, 2].

Find m such that the function y= Aex satisfies the equation 2y" = y + y