Suppose that Newton's method is used to locate a root of the equation f (x) = 0 with initial approximation x1 = 2. If the second approximation is found to be x2 = −6, and the tangent line to f (x) at x = 2 passes through the point (16, 2), find f (2).
+33659 Newton's method says that the root of a function f(x) can be obtained numerically by repeated use of the following formula:
xn = xn-1 - f(xn-1)/f`(xn-1) given an initial guess x1
xn is the n'th estimate; f(xn-1) is the value of the function at xn-1; f`(xn-1) is the value of the gradient (slope/tangent/derivative) at xn-1.
Here, we are told that x1 = 2; x2 = -6; and, because the tangent line at x=16 is 2, we can deduce that
f`(2) = (2 - 0)/(16 - (-6)) = 2/22 = 1/11. So, using the expression above:
x2 = x1 - f(x1)/f`(x1)
-6 = 2 - f(2)/(1/11)
-6 = 2 - f(2)*11
f(2) = 8/11
Same as Chris.
+129850 The tangent line passes through the points [-6, 0], [2, f(2)] and [16, 2]
Equating slopes, we have
[2 - 0] / [16 - (-6) ] = [2 - f(2)] / [16 - 2] simplify
2 / 22 = [2 - f(2)] / 14 simplify
1 /11 = [2 - f(2)] / 14 cross-multiply
14 = 11 [ 2 - f(2)] simplify again
14 = 22 - 11f(2) subtract 22 from each side
-8 = - 11f(2) divide both sides by - 11
8/11 = f(2)
I'm pretty sure this is correct....but since I haven't looked at Newton's Method in a while, I'm going to have Alan look this over.....!!!!
+33659 Newton's method says that the root of a function f(x) can be obtained numerically by repeated use of the following formula:
xn = xn-1 - f(xn-1)/f`(xn-1) given an initial guess x1
xn is the n'th estimate; f(xn-1) is the value of the function at xn-1; f`(xn-1) is the value of the gradient (slope/tangent/derivative) at xn-1.
Here, we are told that x1 = 2; x2 = -6; and, because the tangent line at x=16 is 2, we can deduce that
f`(2) = (2 - 0)/(16 - (-6)) = 2/22 = 1/11. So, using the expression above:
x2 = x1 - f(x1)/f`(x1)
-6 = 2 - f(2)/(1/11)
-6 = 2 - f(2)*11
f(2) = 8/11
Same as Chris.
