Problem 4 - Calculus Early Transcendentals

Question

Give the formula for Newton's method for the function \(f(x)=x^{2}-5\).

Step-by-Step Solution

Answer

Answer: The formula for Newton's Method for the function \(f(x) = x^2 - 5\) is \(x_{n+1} = x_n - \frac{1}{2}(x_n - \frac{5}{x_n})\).

Step 1: Find the derivative of the given function

To find the derivative of the function \(f(x) = x^2 - 5\), we use the power rule for differentiation, which states that \((x^n)' = nx^{n-1}\). So, \(f'(x) = (x^2 - 5)' = (x^2)' - (5)' = 2x - 0 = 2x\).

Step 2: Plug the derivative into the Newton's Method formula

Next, we plug in the values of the function \(f(x)\) and its derivative \(f'(x)\) into the formula for Newton's Method: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) We have \(f(x) = x^2 - 5\) and \(f'(x) = 2x\). Thus, we substitute these into the Newton's Method formula to get \(x_{n+1} = x_n - \frac{(x_n^2 - 5)}{2x_n}\).

Step 3: Simplify the formula

Now, we simplify the formula we derived in step 1: \(x_{n+1} = x_n - \frac{x_n^2 - 5}{2x_n} = x_n - \frac{1}{2} (\frac{x_n^2 - 5}{x_n}) = x_n - \frac{1}{2}(x_n - \frac{5}{x_n})\) So, the formula for Newton's Method for the function \(f(x) = x^2 - 5\) is: \(x_{n+1} = x_n - \frac{1}{2}(x_n - \frac{5}{x_n})\)