91Ó°ÊÓ

Polynomial expansion is the process of expanding an expression or function from a compact form into a more extended form. For the function given in the exercise, we had an expression that involved multiplying a square: \( f(x) = x^2(x-4)^2 \). This means we need to multiply the term \((x-4)\) by itself to expand the polynomial. When we expanded \((x-4)^2\), we applied the distributive property:

• First, multiply \(x-4\) by \(x-4\) to get \(x^2 - 8x + 16\).
• Then, multiply the result by \(x^2\) to transform it into \(x^2(x^2 - 8x + 16)\).

Resulting in a fully expanded polynomial \(x^4 - 8x^3 + 16x^2\). This form is much easier to differentiate, which is an essential next step in finding critical numbers.

Derivative calculation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative, often denoted as \(f'(x)\), represents the rate of change of a function concerning its variable. In our case, we start with the expanded polynomial \(f(x) = x^4 - 8x^3 + 16x^2\).

• Differentiate each term individually:
  • The derivative of \(x^4\) is \(4x^3\).
  • The derivative of \(-8x^3\) is \(-24x^2\).
  • The derivative of \(16x^2\) is \(32x\).
• Combine them to find the full derivative: \(f'(x) = 4x^3 - 24x^2 + 32x\).

This derivative allows us to find where the function is changing—an essential part of locating critical points.

Quadratic factoring is an algebraic method to simplify a quadratic expression by expressing it as a product of its factors. In the derivative, \(f'(x) = 4x^3 - 24x^2 + 32x\), the first step in simplifying is to factor out the greatest common factor (GCF), which is \(4x\). Doing this gives us the expression \(4x(x^2 - 6x + 8)\). Next, we apply quadratic factoring techniques to \(x^2 - 6x + 8\). To factor this expression, we look for two numbers that both multiply to \(+8\) (the constant term) and add up to \(-6\) (the linear coefficient). These numbers are \(-2\) and \(-4\). Thus, \((x^2 - 6x + 8)\) can be factored into \((x-2)(x-4)\). Combining the factored parts, we have \(4x(x-2)(x-4) = 0\), an expression that is much simpler and ready for finding solutions for critical numbers.

Critical points of a function occur where its derivative is equal to zero or undefined. These points can indicate where a function might have a local maximum, local minimum, or point of inflection. From our factored derivative equation \(4x(x-2)(x-4) = 0\), we can determine the critical points.

• Set each factor in the expression equal to zero:
• \(4x = 0\) implies \(x = 0\).
• \(x-2 = 0\) implies \(x = 2\).
• \(x-4 = 0\) implies \(x = 4\).

The critical points are \(x = 0\), \(x = 2\), and \(x = 4\). Each of these values indicates where the function changes behavior, making them potential locations for local extrema or other points where the function's rate of change is significant.