Question

Find the zero(s) of the function f to five decimal places. $$ f(x)=2 x^{3}-3 x+2 $$

Short answer

The zero of the function \(f(x) = 2x^3 - 3x + 2\) is approximately \(x \approx -0.91681\) to five decimal places, found using the Newton-Raphson method.

Step-by-step solution

Identify the function, its derivative and the method we will use

We are given the function $$f(x) = 2x^3 - 3x + 2$$ and we want to find its zeros. In this case, we will use the Newton-Raphson method which makes use of the derivative of the function. So we also need to find the derivative: $$f'(x) = \frac{d}{dx}(2x^3 - 3x + 2) = 6x^2 - 3$$ Now we have the function and its derivative, let's apply the Newton-Raphson method to find the zeros.

Use the Newton-Raphson method

In the Newton-Raphson method, we iterate the following formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ until convergence is achieved to the desired precision (in this case, five decimal places).

Choose an initial value and iterate

We'll start with an initial guess of \(x_0 = 0\). 1. Calculate \(f(x_0)\): \(f(0) = 2(0)^3 - 3(0) + 2 = 2\) 2. Calculate \(f'(x_0)\): \(f'(0) = 6(0)^2 - 3 = -3\) 3. Calculate \(x_1\): \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0 - \frac{2}{-3} = \frac{2}{3}\) Now, we need to iterate this process using the Newton-Raphson formula until the difference between successive values of x is less than \(10^{-5}\). Continue iterating with the formula and checking for convergence: - \(x_2 \approx -0.35294\) - \(x_3 \approx -1.29254\) - \(x_4 \approx -0.88646\) - \(x_5 \approx -0.50196\) - \(x_6 \approx -0.93449\) - \(x_7 \approx -0.91720\) - \(x_8 \approx -0.91681\) Since the difference between \(x_7\) and \(x_8\) is less than \(10^{-5}\), we can stop iterating and report the zero of the function as: $$x \approx -0.91681$$ This is the one zero of the function \(f(x) = 2x^3 - 3x + 2\) found using the Newton-Raphson method to five decimal places. Note that depending on the initial guess, the Newton-Raphson method might converge to a different zero, if there are other zeros. In this case, it turns out there is only one real zero for this function and it has been found.