91Ó°ÊÓ

In calculus, the derivative of a function gives us essential information about its rate of change. Calculating the first derivative of a function is often the initial step in identifying extreme points. For the function given, \( h(x) = x^3 - 8x^2 - 6x \), the power rule is applied to each term:

• The derivative of \( x^3 \) is \( 3x^2 \).
• The derivative of \( -8x^2 \) is \( -16x \).
• The derivative of \( -6x \) is \(-6 \).

Putting these derivatives together, the first derivative is calculated as \( h'(x) = 3x^2 - 16x - 6 \). With this derivative, we are equipped to find critical points, as well as understand the behavior of the function more precisely.

Critical points of a function occur where its derivative equals zero or is undefined. These are potential locations for extrema, or extreme values. In our exercise, we solve the equation \( 3x^2 - 16x - 6 = 0 \) to find critical points. By setting the first derivative to zero, we identify places where the slope of the tangent is horizontal, indicating possible maxima or minima. In this case, the resulting quadratic equation has solutions found using the quadratic formula. These solutions represent the critical points prevalent in the interval being analyzed, \( x = 5.49 \) and \( x = -0.36 \). These points, along with endpoints, are crucial for discovering extreme values within that interval.

The quadratic formula is a standard tool for solving second-degree polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]This calculation involves:

• Identifying the coefficients \( a = 3 \), \( b = -16 \), and \( c = -6 \).
• Substituting these values into the formula.
• Simplifying to find the roots or solutions of the equation.

For our function's derivative, using the quadratic formula gives us the critical points at \( x = 5.49 \) and \( x = -0.36 \). These are vital in ascertaining where the function's rate of change shifts, potentially marking locales of maximum or minimum values.

Once we have the critical points and endpoints of an interval, evaluating the original function at these points helps determine the location of absolute extremes. For the function \( h(x) = x^3 - 8x^2 - 6x \), we probe the critical points and the endpoints \( x = -2 \) and \( x = 10 \). By plugging these values into the function, we discover:

• \( h(-2) = 44 \)
• \( h(5.49) \approx -74.02 \)
• \( h(-0.36) \approx 8.14 \)
• \( h(10) = -260 \)

These evaluations reveal that the function achieves an absolute maximum of 44 at \( x = -2 \) and an absolute minimum of -260 at \( x = 10 \) for the given interval. This step is essential for completing the task of identifying absolute extreme values within a given range.