Numerical Methods Solution Manual: First Course In

Using Lagrange interpolation, we can write the approximate value of f(x) as:

The bisection method involves finding an interval [a, b] such that f(a) and f(b) have opposite signs. In this case, we can choose a = 2 and b = 3, since f(2) = -1 and f(3) = 16. The midpoint of the interval is c = (2 + 3)/2 = 2.5. Evaluating f(c) = f(2.5) = 3.375, we see that f(2) < 0 and f(2.5) > 0, so the root lies in the interval [2, 2.5]. Repeating the process, we find that the root is approximately 2.094568121971209. First Course In Numerical Methods Solution Manual

Evaluating these expressions at x = 0.5, we get: Using Lagrange interpolation, we can write the approximate

where L0(x) = (x - 1)(x - 2)/((0 - 1)(0 - 2)) = (x^2 - 3x + 2)/2, L1(x) = (x - 0)(x - 2)/((1 - 0)(1 - 2)) = -(x^2 - 2x), L2(x) = (x - 0)(x - 1)/((2 - 0)(2 - 1)) = (x^2 - x)/2. Evaluating f(c) = f(2

L0(0.5) = 0.375, L1(0.5) = -0.25, L2(0.5) = 0.0625.