91Ó°ÊÓ

In calculus, finding the derivative of a function is crucial to understanding its behavior. The derivative tells us how a function changes at any given point. For the function \[ f(x) = 8x^3 - 24x^2 + 18x + 6 \]we need to find its derivative to identify critical points. To do this, we use the power rule. The power rule states that the derivative of \( ax^n \) is \( nax^{n-1} \). Applying this to each term:

• The derivative of \( 8x^3 \) is \( 24x^2 \).
• The derivative of \(-24x^2 \) is \(-48x \).
• The derivative of \(18x \) is \( 18 \).
• The derivative of the constant \( 6 \) is \( 0 \).

Bringing these together, the first derivative is:\[ f'(x) = 24x^2 - 48x + 18 \]This derivative will help us find where the function is changing its direction.

To determine where the function is increasing or decreasing, we look at the first derivative:\[ f'(x) = 24x^2 - 48x + 18 \]Critical points occur where this derivative is zero or undefined. For our function, we solve the equation:\[ 24x^2 - 48x + 18 = 0 \]Simplifying, it becomes:\[ 4x^2 - 8x + 3 = 0 \]Using the quadratic formula, we find:\[ x = 0.5 \quad \text{and} \quad x = 1.5 \]These critical points segment the number line into intervals: \((-\infty, 0.5)\), \((0.5, 1.5)\), and \((1.5, \infty)\).

• For \(x < 0.5\), choose \(x = 0\). Plug it into \(f'(x)\), and since the result is positive, the function is increasing.
• For \(0.5 < x < 1.5\), choose \(x = 1\). Plug it into \(f'(x)\), and since the result is negative, the function is decreasing.
• For \(x > 1.5\), choose \(x = 2\). Plug it into \(f'(x)\), and since the result is positive, the function is increasing again.

This gives us an overall picture of where the function rises and falls.

Relative extrema refer to the peaks and valleys in a graph, known as relative maxima and minima. They occur at critical points where the function changes from increasing to decreasing or vice versa.
For \( f(x) = 8x^3 - 24x^2 + 18x + 6 \), we determined critical points at \(x = 0.5\) and \(x = 1.5\) by setting the first derivative to zero.

• At \(x = 0.5\), the function changes from increasing to decreasing, which means there is a relative maximum at \((0.5, f(0.5))\).
• At \(x = 1.5\), the function changes from decreasing to increasing, indicating a relative minimum at \((1.5, f(1.5))\).

To find the exact values of these extrema, evaluate the original function at these points:
\[ f(0.5) = 8(0.5)^3 - 24(0.5)^2 + 18(0.5) + 6 = 4 \]
\[ f(1.5) = 8(1.5)^3 - 24(1.5)^2 + 18(1.5) + 6 = 3 \]
Thus, there is a relative maximum at \((0.5, 4)\) and a relative minimum at \((1.5, 3)\), giving a clear understanding of the function's high and low points.