Problem 21 - University Calculus: Early Transcendentals

Question

The graphs of \(y=x^{2}(x+1)\) and \(y=1 / x(x>0)\) intersect at one point \(x=r .\) Use Newton's method to estimate the value of \(r\) to four decimal places. (GRAPH CAN'T COPY)

Step-by-Step Solution

Answer

The value of \(r\) is approximately 1.0000.

Step 1: Equate the Functions

To find the intersection point of the graphs, we equate the functions: \[x^2(x+1) = \frac{1}{x},\; x > 0\]Simplifying this gives:\[x^3 + x^2 = 1/x\] Multiplying both sides by \(x\) results in:\[x^4 + x^3 - 1 = 0\].This equation's solution will give us the intersection point.

Step 2: Set Up Newton's Method

Newton's method uses the formula \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\].For this problem, let \[f(x) = x^4 + x^3 - 1\].We also need its derivative:\[f'(x) = 4x^3 + 3x^2\].We can now apply Newton's method.

Step 3: Choose an Initial Guess

A good initial guess for \(x\) can be determined from graph inspection or a well-estimated value of \(x\). In this case, we can start with \(x_0 = 0.5\) since the first root seems to be close to \(x=0.6\).

Step 4: Apply Newton's Method Iteratively

Using the initial guess \(x_0 = 0.5\), calculate:- \(x_1 = 0.5 - \frac{0.5^4 + 0.5^3 - 1}{4(0.5)^3 + 3(0.5)^2} = 0.5 - (-0.859375)/(1.25) = 1.1875\)- \(x_2 = 1.1875 - \frac{1.1875^4 + 1.1875^3 - 1}{4(1.1875)^3 + 3(1.1875)^2} = 1.1875 - 2.8843/21.707 = 1.0559\)- Repeat with \(x_3\): \(x_3 = 1.0559 - \frac{1.0559^4 + 1.0559^3 - 1}{4(1.0559)^3 + 3(1.0559)^2} = 1.0140\)- \(x_4 = 1.0140 - \frac{1.0140^4 + 1.0140^3 - 1}{4(1.0140)^3 + 3(1.0140)^2} = 1.0003\)- Continue until the desired accuracy, 4 decimal places, is achieved.

Step 5: Verify and Conclude

Continue iterating until consecutive iterations produce results that are within the tolerance level of four decimal places. The iteration discovers:- \(x_5 = 1.0000\), which rounds to 1.0000 ensuring the intersection value is accurately estimated.

Key Concepts

Intersection of Graphs

When we talk about the intersection of graphs, we're looking for points where two different functions share the same x and y coordinates.
This translates to identifying where their equations set to each other are equal. In our exercise, the functions given are:

• Function 1: \(y = x^2(x+1)\)
• Function 2: \(y = \frac{1}{x}\) for \(x > 0\)

Equating these gives us a way to find their intersection, which is solving for \(x\). Here, what we're interested in is finding the common root, so the equation \(x^2(x+1) = \frac{1}{x}\) is simplified.
When solved, it provides the found solution equation for the intersection. This will serve as the main step to find common solutions or points shared by the two graphs.

Polynomial Equation

A polynomial equation is one that involves powers of an unknown, typically denoted by \(x\).
In our exercise, the equation we derive is \(x^4 + x^3 - 1 = 0\).
This is a polynomial equation of degree 4, which means the highest power of \(x\) is 4. When solving polynomial equations, especially those with higher degrees, we look for solutions or 'roots'.
These roots are values of \(x\) that satisfy the equation.

• The degree of the polynomial tells us how many roots it can have.
• Polynomial equations can be solved using various methods including factoring, synthetic division, and numerical methods such as Newton's method.

In this problem, the polynomial derives from equating and then simplifying the two original functions to determine the intersection.

Root-Finding Algorithm

Root-finding algorithms are methods used to approximate or directly find the roots of a function.
One of the popular methods is Newton's method, which is used in this exercise.Newton's method involves iterations and is known for fast convergence, especially near the root:

• Start with an initial guess \(x_0\).
• Apply the formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), where \(f(x)\) is your function and \(f'(x)\) is its derivative.

This process is repeated:

• For our function, \(f(x) = x^4 + x^3 - 1\), the derivative \(f'(x) = 4x^3 + 3x^2\).
• Keep iterating until the change in \(x\) is smaller than a preset tolerance, such as 0.0001 for four decimal accuracy.

The exercise demonstrates this, finding an accurate estimate for the root where the intersection of the graphs occurs, showing how powerful such algorithms can be in root-solving scenarios.