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Differentiation is a process used to find the rate of change of a function with respect to a variable. It is a core concept in calculus that helps us understand how functions behave. When we differentiate a function, we are essentially determining its derivative. The derivative is a new function that tells us the slope of the original function at any given point.
Here's a brief look at how differentiation works:

• The derivative of a constant is zero since constants don’t change.
• The derivative of a function like \(x^n\) is \(nx^{n-1}\).
• If you have a function with multiple terms, find the derivative of each term separately.

For the function \(f(x) = 3x^3 + \frac{3x^2}{2} - 2x\), differentiating each term with respect to \(x\) gives us \(f'(x) = 9x^2 + 3x - 2\). This new function \(f'(x)\) reflects how \(f(x)\) changes at any point. By setting the derivative equal to zero, we find critical points that help in determining where the function increases or decreases.

The quadratic formula is a powerful method for finding the roots (or solutions) of quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). The general formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula gives solutions by calculating the values of \(x\) that make the quadratic equation true. The steps include:

• Identifying \(a\), \(b\), and \(c\) from the quadratic equation.
• Substituting these values into the formula.
• Calculating the discriminant \(b^2 - 4ac\) to determine the nature of the roots.
• Simplifying to find \(x\).

In our example, \(9x^2 + 3x - 2 = 0\) can be solved using this formula with \(a = 9\), \(b = 3\), and \(c = -2\), resulting in the critical points \(x_1 = \frac{-3 + \sqrt{57}}{18}\) and \(x_2 = \frac{-3 - \sqrt{57}}{18}\). The solutions are the \(x\)-values where the slope of \(f(x)\) is zero, highlighting crucial changes in the behavior of the function.

A local maximum occurs at a point where a function changes direction from increasing to decreasing. At this point, the function reaches the highest value when viewed through a small "local" window. Visualizing this on a graph, it's like a hill's peak. Understanding local maxima helps in identifying optimal solutions and strategizing around turning points.
To find a local maximum through calculus:

• Calculate the derivative of the function to find critical points by setting it to zero.
• Analyze these points using the first derivative test or look at a graph to observe changes.

In our situation, when the function \(f(x) = 3x^3 + \frac{3x^2}{2} - 2x\) is graphed, the critical point \(x_1 = \frac{-3 + \sqrt{57}}{18}\) was identified as a local maximum. Here, the function's graph transitions from an uphill slope to a downhill one, signifying a peak or local maximum.

A local minimum in a function occurs where the function changes from decreasing to increasing, representing the lowest point within a given interval. Imagine this as a valley in the function's graph. Identifying a local minimum can be useful in problem-solving for finding the most efficient or least costly paths.
Steps to locate a local minimum include:

• Compute the function's derivative to locate critical points.
• Determine the nature of each point using the first derivative test or graphical analysis.

In the context of our function \(f(x) = 3x^3 + \frac{3x^2}{2} - 2x\), the critical point \(x_2 = \frac{-3 - \sqrt{57}}{18}\) is identified as a local minimum. The graph confirms this as the slope moves from negative (downgoing) to positive (upgoing). This change indicates a trough or valley, signaling a local minimum.