91Ó°ÊÓ

Derivatives are a fundamental tool in calculus. They measure how a function changes as its input changes. You can think of a derivative as the "rate of change" or the "slope" of a function at any given point. In the context of our exercise, we started by finding the first derivative of the function \( y = x^3 - 12x + 12 \). This step was essential for identifying where the graph changes direction from increasing to decreasing, and vice versa. We calculated the first derivative to be \( y' = 3x^2 - 12 \). Breaking it down, the derivative is like a dynamic lens, zooming into the function's growth behavior. For every term in our polynomial, the power rule was applied. The power rule essentially states that if you have \( x^n \), its derivative is \( nx^{n-1} \). So, \( 3x^2 \) is the derivative of \( x^3 \), and \(-12\) is the derivative of \(-12x\). By setting \( y' = 0 \), we could find critical points where the function potentially has a maximum or minimum.

Critical points in calculus help us determine where a function's graph has peaks or valleys. These are points where the derivative equals zero or is undefined. In our problem, after calculating the first derivative \( 3x^2 - 12 \), we set it to zero to find these points. Solving the equation \( 3(x^2 - 4) = 0 \) led us to the critical points \( x = 2 \) and \( x = -2 \). These values tell us that something significant happens to the function's slope at these inputs, namely that there could be a local maximum or minimum at each point.
It's important to consider that finding a critical point is like knowing where the mountain tops and valleys of a graph could potentially be, but we don't know exactly what kind they are—that's where checking the function's concavity comes into play.

Concavity describes how the slope of a function changes—not just the direction of change, but the nature of that change. Think of it as how a graph "curves." We use the second derivative to assess concavity, which involves differentiating the first derivative—this led us to \( y'' = 6x \). By substituting our critical points into this second derivative, we can determine whether those points are local maxima or minima for the graph.

• At \( x = 2 \), \( y''(2) = 12 \): the positive result indicates the graph is concave up, akin to a bowl. Hence, \( x = 2 \) is a local minimum.
• At \( x = -2 \), \( y''(-2) = -12 \): the negative result indicates the graph is concave down, similar to a hilltop. Therefore, \( x = -2 \) is a local maximum.

This concavity test is crucial for confirming the nature of each critical point.

Graph sketching ties together everything we've determined: the derivatives, critical points, and concavity. The purpose of sketching the graph is to provide a visual understanding of a function's behavior.
For our function \( y = x^3 - 12x + 12 \), the critical points \( x = 2 \) and \( x = -2 \) play a significant role. We've established:

• A local maximum at \( x = -2 \).
• A local minimum at \( x = 2 \).

Additionally, considering the behavior of the graph as \( x \to \pm \infty \), the \( x^3 \) term dominates, causing \( y \to \infty \). These insights help sketch the general shape of the graph. Always remember, sketching is an estimation. Use technology, like a graphing calculator, to verify and refine your sketch for precision.