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The Newton-Raphson method is an iterative method used to find the approximate roots of a function. The method starts with an initial guess \(x_0\) and refines the guess using the iterative formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) We will use this formula to find the zeros of the given function to five decimal places.

Before we can use the Newton-Raphson method, we need to calculate the first derivative of the function. For our function \( f(x) = x^3 - 9x + 4 \), the first derivative is: \( f'(x) = \frac{d}{dx} (x^3 - 9x + 4) = 3x^2 - 9 \)

We need to start with an initial guess for the zero of the function. We can use trial and error, or we can plot the function and see where it appears to be close to zero. Based on either of these methods, we will guess that one zero of the function is near \(x = 2\). So, we will set \( x_0 = 2 \).

Now that we have an initial guess, we can apply the Newton-Raphson method to refine our guess. Using the formula from Step 1, we update our guess as follows: 1. Calculate \( f(x_0) = f(2) = 2^3 - 9(2) + 4 = -8 \) and \( f'(x_0) = f'(2) = 3(2)^2 - 9 = 3 \) 2. Update \( x_0 \) using the Newton-Raphson formula: \( x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-8}{3} = 2 + \frac{8}{3} \approx 4.67 \) We will repeat these steps until the difference between two consecutive guesses is less than \(10^{-5}\). Continue updating the guesses using the Newton-Raphson formula: Eventually, we get \( x_4 \approx 3.61155 \) and \( x_5 \approx 3.61154 \), since the difference between them is less than \(10^{-5}\), we can round our answer to five decimal places and say that one of the zeros is \(x \approx 3.61154\).

We've found one zero of the function, but we should check for additional zeros. We could apply the Newton-Raphson method again with new initial guesses, or we could notice that this function appears to have only one zero (based on the graph or additional analysis). In this case, there is only one real zero, so our final answer is: \(x \approx 3.61154\).