Question

More root finding Find all the roots of the following functions. Use preliminary analysis and graphing to determine good initial approximations. $$f(x)=x^{2}(x-100)+1$$

Short answer

Answer: To find the roots of a function, we can follow these general steps: 1. Analyze the function and find critical points using its derivative for initial approximations. 2. Apply numerical methods, such as the Newton-Raphson method, with the initial approximations to refine the values of the roots iteratively. For the given function, $$f(x) = x^2(x - 100) + 1$$, we first find the critical points at x = 0 and x ≈ 66.67. We then use the Newton-Raphson method with initial approximations around these critical points (e.g., x = 30 and x = 80) to iteratively refine the values of the roots until the desired level of accuracy is achieved.

Step-by-step solution

Preliminary analysis and graphing

To begin, let's analyze the given function: $$f(x) = x^2(x - 100) + 1$$ First, let's find the critical points, which are the points where the derivative of the function is equal to zero. To do this, we need to differentiate the function with respect to x. Let's use the product rule and find the derivative: $$f'(x) = d[(x^2)(x-100)]/dx + 0$$ $$f'(x) = (x^2) (1) + (x-100) (2x)$$ $$f'(x) = x^2 + 2x(x - 100)$$ $$f'(x) = x^2 + 2x^2 - 200x$$ $$f'(x) = 3x^2 - 200x$$ Now, let's find when the derivative equals zero: $$3x^2 - 200x = 0$$ $$x(3x - 200) = 0$$ Thus, the critical points are x = 0 and x = 200/3 ≈ 66.67. By examining the graph of the function or finding the second derivative, we can see that the function has a local minimum at x = 0 and a local maximum at x ≈ 66.67. Therefore, there might be one root between these two critical points and another root beyond x ≈ 66.67. Now let's use numerical methods to refine these initial approximations to find accurate roots.

Using numerical methods for root finding

We will now use the Newton-Raphson method for root finding. This method requires an initial approximation and iteratively refines it to find the root. The Newton-Raphson formula is given by: $$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$$ In our case, we already have the function and its derivative: $$f(x) = x^2(x - 100) + 1$$ $$f'(x) = 3x^2 - 200x$$ Let's first refine the approximation for the root between the critical points of x = 0 and 66.67. Initial approximation: $$x_0 = 30$$ (midpoint between the critical points) Now, run several iterations of the Newton-Raphson formula: $$x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)}$$ $$x_2 = x_1 - \dfrac{f(x_1)}{f'(x_1)}$$ Continue this process until you reach an acceptable level of accuracy in the value of the root. Next, let's refine the approximation for the root beyond x ≈ 66.67. Initial approximation: $$x_0 = 80$$ Repeat the above iteration process using the Newton-Raphson formula to obtain a refined value of the second root. After running several iterations of the Newton-Raphson method, the refined roots of the given function can be obtained.

Key concepts

Critical Points

In calculus, critical points are essential in understanding the behavior of functions. Critical points occur where the derivative of a function equals zero or where the derivative does not exist.

For the function
\( f(x) = x^2(x - 100) + 1 \),
finding critical points involves taking the derivative and setting it equal to zero. This reveals where the function's slope is flat, indicating potential local maxima, minima, or inflection points. In the given exercise, critical points provided initial guesses for root finding, and they are crucial in analyzing a function's overall pattern.

Newton-Raphson Method

The Newton-Raphson method is a powerful technique for finding numerical approximations to the roots (or zeroes) of a real-valued function. It is an iterative method that starts with an initial estimate and employs the function's derivative

to refine this estimate in a systematic way. Specifically,

\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
represents the iterative formula. The choice of initial approximation (\(x_0\)) is important, as a poor choice can lead to divergence from the actual root. In practice, this method is highly efficient and often converges to an accurate root swiftly if the initial guess is reasonably close.

Derivatives

In our context, derivatives reflect the rate of change of a function with respect to its variable. It gives the slope of the tangent line to the curve of the function at any given point.

When we found the derivative of our function,\( f'(x) = 3x^2 - 200x \), we used it to identify critical points for our function and also within the Newton-Raphson method to refine our estimate of the roots. Derivatives are fundamental to both aspects of our problem-solving strategy, as they provide insights into the function's behavior and are integral to the root refinement process.

Product Rule

The product rule is a derivative rule used when differentiating products of two or more functions. It states that the derivative of a product is the derivative of the first function multiplied by the second function, plus the first function times the derivative of the second function.

If we have functions \(u(x)\) and \(v(x)\), then their derivative using the product rule is given by:

\[ \frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]
In our exercise solution, we applied the product rule to differentiate \(f(x) = x^2(x - 100)\), which allowed us to find \(f'(x)\) crucial for identification of critical points and implementing the Newton-Raphson method. Understanding and applying the product rule is crucial for solving calculus problems involving the derivatives of product functions.