Use Newton's method to approximate the indicated root of the equation correct to six decimal places. The root of x4 − 2x3 + 7x2 − 9 = 0 in the interval [1, 2]

Answer 1

Newton's Method uses the following equation to iterate closer and closer to the correct answer:

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

Notice that this number is the same as the previous iteration so this is the value that is accurate to 9 decimal places. Since we are instructed to get an answer accurate to 6 decimal places, we just round to 6 so the answer is 1.219842

This whole thing might seem tedious but if you have a calculator like the TI-84, you can use it to make this easy. Type in the function into Y₁ and the derivative into Y₂ then use your function value button (either use the sequence: "vars, Y-VARS, Function" or use the shortcut "alpha-trace" to select either Y₁ or Y₂. For your first iteration you would type in your calculator 1.5-Y₁(1.5)/Y₂(1.5) and you would get 1.258928571