Question 12.10: An approximate root of x³ - 2x² - 5 = 0 is x = 3. By using......
An approximate root of
x³ – 2x² – 5 = 0
is x = 3. By using the Newton-Raphson technique repeatedly, determine the value of the root correct to two decimal places.
Step-by-Step
We have
\begin{array}{ll} x_1=3 & \\ f(x)=x^3-2 x^2-5 & f\left(x_1\right)=4 \\ f^{\prime}(x)=3 x^2-4 x & f^{\prime}\left(x_1\right)=15 \end{array}Hence
x_2=3-\frac{4}{15}=2.733An improved estimate of the value of the root is 2.73 (2 d.p.). The method is used again, taking x_1 = 2.73 as the initial approximation:
\begin{aligned} x_1 & =2.73 \quad f\left(x_1\right)=0.441 \quad f^{\prime}\left(x_1\right)=11.439 \\ x_2 & =2.73-\frac{0.441}{11.439}=2.691 \end{aligned}An improved estimate is x = 2.69 (2 d.p.). The method is used again:
x_1=2.69 \quad f\left(x_1\right)=-0.007 \quad f^{\prime}\left(x_1\right)=10.948So
x_2=2.69-\frac{(-0.007)}{10.948}=2.691There is no change in the value of the approximate root and so to two decimal places the root of f(x) = 0 is x = 2.69.
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