91Ó°ÊÓ

Let's start with understanding a cubic equation. A cubic equation is a polynomial of degree three. It has the general form:
\[\begin{equation}ax^3 + bx^2 + cx + d = 0ewline\text{where } a eq 0ewline\text{These equations can have one, two, or three real roots.}ewline\text{In our exercise, the equation } x^3 - 12x - 16 = 0 \text{ is a cubic equation.} ewline\end{equation}\] To solve a cubic equation, you can use several methods. Popular techniques include:

• Trial and Error
• Rational Root Theorem
• Graphical Methods
• Cardano's Formula

Let's use the rational root theorem quickly. It suggests that all possible roots of a polynomial are the factors of the constant term divided by the factors of the leading coefficient. In our equation, the possible rational roots are the factors of -16 divided by the factors of 1, which gives us: ±1, ±2, ±4, ±8, and ±16.
By testing these, we determine that one solution is x=4.

Roots of an equation are the values of the variable that satisfy the equation. For polynomial equations, the roots denote where the polynomial equals zero. In our cubic equation \[\begin{equation} x^3 - 12x - 16 = 0 ewline, x=4 ewline is a root. ewline\text{You can verify this by substituting } x=4 text { back into the equation:} ewline 4^3 - 12 \times 4 - 16 \rightarrow 64 - 48 - 16 = 0 \text{, which is true.}\end{equation}\] Verifying roots in this way ensures there's no mistake. Root-finding is crucial in defining critical points in graphs. In this exercise, finding that x=4 is a root implies it's a point where the curve intersects or touches the x-axis.

Implicit differentiation involves finding the derivative of a function without solving for one variable in terms of the other. This technique is essential when dealing with equations where y cannot be easily isolated.
Given the original curve equation \[\begin{equation} x^3 + y^3 - 12x - 8y - 16 = 0 \text{, applying implicit differentiation yields}ewline 3x^2 + 3y^2\frac{dy}{dx} - 12 - 8\frac{dy}{dx} = 0ewline\text{Rewriting and simplifying it at y=0 gives}ewline 3x^2 - 12 = 0ewline 3x^2 = 12ewline x^2 = 4ewline x=\text{±}2 \text{.}\end{equation}\] At this point, we see that when x=4, the slope, given by \frac{dy}{dx}, equals zero, confirming the curve touches the x-axis. Using implicit differentiation helps check tangencies or slopes when straightforward differentiation isn't feasible.