Question

A root of \(x^{3}-3 x^{2}+2.5=0\) lies between \(1.1\) and \(1.2\).

Short answer

A root is located between 1.1 and 1.2.

Step-by-step solution

Understand the Problem

We need to find a root of the cubic equation \(x^3 - 3x^2 + 2.5 = 0\) between \(x = 1.1\) and \(x = 1.2\). To do this, we will evaluate the function at these points to check where the sign changes, which indicates the presence of a root.

Evaluate the Function at x = 1.1

Substitute \(x = 1.1\) into the equation:\[f(1.1) = (1.1)^3 - 3(1.1)^2 + 2.5\]Calculate this value:\[f(1.1) = 1.331 - 3 imes 1.21 + 2.5 = 1.331 - 3.63 + 2.5 = 0.201\]Since \(f(1.1) > 0\), the function is positive at \(x = 1.1\).

Evaluate the Function at x = 1.2

Now, substitute \(x = 1.2\) into the equation:\[f(1.2) = (1.2)^3 - 3(1.2)^2 + 2.5\]Calculate this value:\[f(1.2) = 1.728 - 3 imes 1.44 + 2.5 = 1.728 - 4.32 + 2.5 = -0.092\]Since \(f(1.2) < 0\), the function is negative at \(x = 1.2\).

Analyze the Results

We have \(f(1.1) = 0.201\) and \(f(1.2) = -0.092\). The sign changes from positive to negative between \(x = 1.1\) and \(x = 1.2\). This indicates that there is a root between these two values, according to the Intermediate Value Theorem.

Key concepts

Cubic Equations

Cubic equations are polynomial equations of degree three. The general form of a cubic equation is given by:

• \[f(x) = ax^3 + bx^2 + cx + d = 0\]

where \(a\), \(b\), \(c\), and \(d\) are constants, and \(a eq 0\). In our specific example, the equation is \(x^3 - 3x^2 + 2.5 = 0\). This equation will have three roots, which could be real or complex.
Understanding how to find these roots is crucial to solving cubic equations. The roots are the values of \(x\) that make the equation equal to zero, and finding them requires various techniques.

Depending on the specific equation, these techniques can include:

• Factoring
• Using the Rational Root Theorem
• Graphical methods
• Numerical methods such as the Newton-Raphson method

However, to find specific intervals where these roots exist, the Intermediate Value Theorem is often used as a preliminary step.

Root Finding

Root finding is a process of determining the values of \(x\) for which a given function, like our cubic equation, becomes zero. Specifically, for the equation \(x^3 - 3x^2 + 2.5 = 0\), we need to determine where the function crosses the x-axis. This tells us about the real roots of the equation.

There are several methods available to locate the roots:

• Analytical methods: Simplifying or factoring the expression to extract roots directly when possible. For more complex equations, this might not be practical without techniques like the quadratic formula or completing the square.
• Graphical methods: Plotting the function to visually identify where it intersects the x-axis. This can give a rough idea of where the roots are located.
• Sign Change Method: This involves evaluating the function at various points to detect sign changes, which signal the presence of a root between those points.
• Numerical methods: Such as the Newton-Raphson method, which can accurately approximate roots through iterative calculations starting from an initial guess.

In the exercise, to confirm the root is between 1.1 and 1.2, we rely primarily on the Sign Change Method for validation.

Sign Change Method

The sign change method is a helpful technique for finding roots, especially when dealing with continuous functions like polynomials. By evaluating a function at specific points, this method identifies intervals where the sign of the function changes, signaling that a root exists within that interval.

Here's how it works in practice:

• Calculate the function value at several points within a given domain.
• Observe where the function's value goes from positive to negative or vice versa.
• Pinpoint these intervals to determine where roots might be located.

In our example, we analyzed the cubic equation \(x^3 - 3x^2 + 2.5\) by evaluating at \(x = 1.1\) and \(x = 1.2\). The calculations showed:

• \(f(1.1) = 0.201\) which is positive.
• \(f(1.2) = -0.092\) which is negative.

The positive-to-negative transition between these points suggests a root exists between 1.1 and 1.2. This application of the Intermediate Value Theorem gives us a reliable indication of the root's presence within that interval.