The indicated root of the equation correct to six decimal places is 1.219841771
Newton's Method uses the following equation
Xn+1 = Xn - (f'(xn)/f'(xn))
To use this, we must first calculate the derivative f '(x). Use the power rule to do it.
f '(x) = 4x3 - 6x2 + 14x
For f(x), use f(x) = x4 - 2x3 + 7x2 - 9
Since we are told to use the interval [1, 2], I'll arbitrarily pick x0 = 1.5 to start with. So the equation to estimate x1 is as follows:
x1 = x0 - f(x0)/f '(x0)
x0 = 1.5
f(x0) = f(1.5) = (1.5)4 − 2(1.5)3 + 7(1.5)2 − 9 = 5.0625
f '(x0) = f '(1.5) = 4(1.5)3 - 6(1.5)2 + 14(1.5) = 21
So x1 = 1.5 - 5.0625/21 = 1.258928571
Repeat this process using x1 = 1.258928571
x2 = x1 - f(x1)/f '(x1)
x2 = 1.258928571 - f(1.258928571)/f '(1.258928571) = 1.258928571 - 0.6156650015/16.09670246 = 1.220680675
x3 = x2 - f(x2)/f '(x2)
x3 = 1.220680675 - f(1.220680675)/f '(1.220680675) = 1.220680675 - 0.0129337886/15.42471777 = 1.219842164
x4 = x3 - f(x3)/f '(x3)
x4 = 1.219842164 - f(1.219842164)/f '(1.219842164) = 1.219842164 - 6.049487E-6/15.41027415 = 1.219841771
x5 = x4 - f(x4)/f '(x4)
x5 = 1.219841771 - f(1.219841771)/f '(1.219841771) = 1.219841771 - (-6.7489E-9)/15.41026739 = 1.219841771