91Ó°ÊÓ

Polynomial functions are algebraic expressions that consist of variables raised to whole-number exponents and coefficients. These functions are foundational in mathematics and have wide applications. Consider the polynomial function from the exercise:

• This function is given as: \[ y = x^4 + 5x^3 - 6x^2 + 7x - 3 \]
• Each term consists of:
  • a coefficient (e.g., 5 in \(5x^3\))
  • and a variable raised to an exponent (e.g., \(x^3\)).

The degree of the polynomial is determined by the highest power of the variable. Here, the degree is 4 because of the term \(x^4\). Understanding the structure of polynomial functions is essential for differentiation. Recognizing each component aids in deriving expressions such as the one needed in this exercise.

To find the derivative of a polynomial function like the one in this exercise, we apply the power rule. This rule states that:

• The derivative of \(x^n\) is \(nx^{n-1}\).

Using this rule, we differentiate each term in the polynomial function.The derivative calculation steps for our specific example include:

• The derivative of \(x^4\) is \(4x^3\).
• The derivative of \(5x^3\) is \(15x^2\).
• The derivative of \(-6x^2\) is \(-12x\).
• The derivative of \(7x\) is \(7\).
• The constant \(-3\) differentiates to \(0\).

Combining these results, the expression for the derivative is:\[ \frac{\mathrm{d} y}{\mathrm{d} x} = 4x^3 + 15x^2 - 12x + 7 \] Understanding how to differentiate each term separately and then combine them can simplify complex calculations and help comprehend the behavior of the function across the graph.

In mathematics, the derivative is often understood as a tool to determine the rate of change. Specifically, it provides the rate at which one quantity changes in relation to another.

• In the context of our exercise, it describes how the polynomial \(y\) changes as \(x\) varies.

For practical purposes, evaluating the derivative at a specific point, in this case, \(x = -2\), reveals how the function behaves at that point. By substituting \(x = -2\) into the differentiated expression:\[ \frac{\mathrm{d} y}{\mathrm{d} x} = 4(-2)^3 + 15(-2)^2 - 12(-2) + 7 = 59 \] This value of 59 indicates that at \(x = -2\), the function \(y\) is increasing at a rate of 59 units per unit increase in \(x\). Understanding this concept is crucial, as it translates the abstract differentiation into a tangible measure, providing insight into the function's instantaneous rate of change and helping predict behavior near the evaluated point.