Newton’s Method is an efficient method for finding the x-intercepts(or real zeros) of a function, such as p(x). step 1:
select an initial value x(lower 0) that is somewhat close to the x-intercept being sought. step 2: find the values for x
using the relation x x = x - p(x lower n ) / p'(x lower n) n+1 n <<< n=1,2.... until you get two consecutive values
x(lower k ) and x(lower k+1) that agree to whatever decimal place accuracy you desire. step 3: the approximate zero
will be x(lower case k+1) consider the polynomial p(x)=x^3 - 7x - 40. a) evaluate p(5) and p(3) b) what might we
conclude about a zero of p. explain. c) use Newton’s method to approximate and x-intercept, r, 3
p(x) = x^3 - 7x - 40; p’(x) = 3x^2 - 7
p(5) = 5^3 - 7(5) - 40 = 50 and p(3) = (3)^3 - 7(3) - 40 = -34
Since p(5) is positive and p(3) is negative, this means p(x) has a root between x = 5 and x = 3
Let us take the starting (trial) value for Newton's method as x0 = 4
x1 = x0 - p(x0)/p’(x0) = 4 - p(4)/p’(4) = 4 - (-4/41) = 4.0975
x2 = x1 - p(x1)/p’(x1) = 4.0975 - p(4.0975)/p’(4.0975) = 4.0949
x3 = x2 – p(x2)/p’(x2) = 4.0949 – p(4.0949)/p’(4.0949) = 4.0949
Since we got two repeating values, the root of p(x) correct to 4 decimal places is x = 4.0949.