Given And For Which Value Of Does

Given and For Which Value of Does

This article delves into the mathematical problem: "Given f(x) = x^3 - 6x^2 + 11x - 6, for which value of x does f(x) = 0?" We will explore multiple methods to solve this cubic equation, emphasizing the importance of understanding the underlying concepts and techniques. We'll also discuss the broader context of solving polynomial equations and their applications.

Understanding Cubic Equations

A cubic equation is a polynomial equation of degree three. The general form is represented as:

ax³ + bx² + cx + d = 0

where 'a', 'b', 'c', and 'd' are constants, and 'a' is not equal to zero. Our specific equation, f(x) = x³ - 6x² + 11x - 6 = 0, fits this form with a=1, b=-6, c=11, and d=-6. Solving this equation means finding the values of 'x' that make the equation true, i.e., the roots or zeros of the function.

Method 1: Factoring the Cubic Equation

Factoring is often the most straightforward method for solving cubic equations, especially when the roots are integers or rational numbers. The goal is to express the cubic polynomial as a product of linear factors. For our equation:

x³ - 6x² + 11x - 6 = 0

We can try to find integer roots using the Rational Root Theorem. This theorem states that any rational root of the equation will be of the form p/q, where p is a factor of the constant term (d = -6) and q is a factor of the leading coefficient (a = 1).

Therefore, potential rational roots are ±1, ±2, ±3, and ±6. We can test these values by substituting them into the equation:

• Testing x = 1: 1³ - 6(1)² + 11(1) - 6 = 0. Therefore, x = 1 is a root.

Since x = 1 is a root, (x - 1) is a factor. We can perform polynomial long division or synthetic division to find the other factor:

(x³ - 6x² + 11x - 6) / (x - 1) = x² - 5x + 6

Now we have a quadratic equation:

x² - 5x + 6 = 0

This quadratic can be easily factored:

(x - 2)(x - 3) = 0

Thus, the roots of the quadratic are x = 2 and x = 3.

Conclusion: The values of x for which f(x) = 0 are x = 1, x = 2, and x = 3.

Method 2: Using the Cubic Formula

The cubic formula, similar to the quadratic formula, provides a general solution for cubic equations. However, it is significantly more complex and often less practical than factoring. The cubic formula is given by:

[Insert Cubic Formula Here - This is a very lengthy formula and including it verbatim would significantly detract from readability. A link to a reliable source showing the formula would be more beneficial].

While powerful, using the cubic formula directly for this specific equation would be unnecessarily cumbersome. The factoring method is far more efficient in this case.

Method 3: Numerical Methods (For More Complex Cases)

When factoring isn't easily achievable, numerical methods such as the Newton-Raphson method or the bisection method can approximate the roots of a cubic equation. These iterative methods refine an initial guess until a root is found to a desired level of accuracy.

Newton-Raphson Method: This method uses the derivative of the function to iteratively improve the approximation of the root. The iterative formula is:

x_(n+1) = x_n - f(x_n) / f'(x_n)

where x_n is the current approximation, f(x_n) is the function value at x_n, and f'(x_n) is the derivative of the function at x_n.

For our function, f'(x) = 3x² - 12x + 11. Starting with an initial guess, the method iteratively refines the approximation until convergence.

Bisection Method: This method repeatedly bisects an interval where a root is known to exist, narrowing down the search until the root is found within a desired tolerance.

These numerical methods are especially valuable when dealing with cubic equations that don't have easily factorable solutions or when dealing with irrational or complex roots.

Applications of Cubic Equations

Cubic equations find applications in various fields:

• Engineering: Designing structures, analyzing fluid flow, and determining optimal shapes.
• Physics: Modeling motion, calculating trajectories, and solving problems in mechanics.
• Chemistry: Determining reaction rates and equilibrium concentrations.
• Economics: Modeling economic growth and analyzing market trends.
• Computer Graphics: Creating curves and surfaces.

Further Exploration: Higher-Degree Polynomial Equations

The techniques discussed for cubic equations can be extended, although with increasing complexity, to higher-degree polynomial equations. While factoring becomes progressively more challenging, numerical methods remain a valuable tool for finding approximate solutions. The fundamental theorem of algebra guarantees that a polynomial of degree 'n' has exactly 'n' roots (counting multiplicity), including real and complex numbers.

Conclusion

Solving the equation x³ - 6x² + 11x - 6 = 0 demonstrates the fundamental principles of solving cubic equations. The factoring method proved highly effective in this case, providing the roots x = 1, x = 2, and x = 3. Understanding these methods and their underlying mathematical concepts is crucial for tackling more complex polynomial equations encountered in various fields of study and application. The choice of method often depends on the specific equation's characteristics and the level of accuracy required. Remember to always check your solutions by substituting them back into the original equation.