Problem 15 - College Algebra and Calculus: An Applied Approach

Question

Describe the increasing and decreasing behavior of the function. Find the point or points where the behavior of the function changes. \(f(x)=x^{3}-3 x^{2}\)

Step-by-Step Solution

Answer

The function \(f(x) = x^{3} - 3x^{2}\) increases on the intervals \(-\infty\) to \(0\) and \(2\) to \(+\infty\), and it decreases on the interval from \(0\) to \(2\).

Step 1: Compute the derivative of the function

The derivative of the function \(f(x) = x^{3} - 3x^{2}\) is computed using the power rule for differentiation, which states that the derivative of \(x^n\) is \(nx^{n-1}\). Hence, the derivative, \(f'(x)\), is \(f'(x) = 3x^{2} - 6x\).

Step 2: Find where the derivative equals zero

The points where \(f'(x) = 0\) are potential points where the function may change from increasing to decreasing or vice versa. Solving the equation \(3x^{2} - 6x = 0\) gives \(x=0\) and \(x=2\). These are our potential critical points.

Step 3: Determine the intervals where the function increases or decreases

This is done by taking a number in each interval defined by the critical points and plugging it into the derivative. If \(f'(x) > 0\) in an interval, the function is increasing. If \( f'(x) < 0\), the function is decreasing. For \(x<0\), \(f'(x) > 0\); between \(0 < x < 2\), \(f'(x) < 0\); for \(x>2\), \( f'(x) > 0\). Hence, the function is increasing from \(-\infty\) to \(0\), decreasing from \(0\) to \(2\), and increasing again from \(2\) to \(+\infty\).

Key Concepts

Calculus

Calculus is a branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series. It's a tool for solving problems in science, engineering, economics, and almost any field that involves a changing system. In the context of understanding function behavior, such as increasing or decreasing trends, calculus is instrumental.

For example, in the exercise with the function \(f(x)=x^{3}-3 x^{2}\), calculus is used to analyze how the function's value changes as the variable \(x\) increases or decreases. Through the process of differentiation, a core component of calculus, we determine the function's rate of change at any point.

Derivatives

Derivatives represent the rate at which a function is changing at any given point. In simpler terms, think of a derivative as the slope of a curve on a graph at a specific point. If you're looking at a hill, the derivative tells you how steep the hill is when you're standing on it. In the exercise given, the derivative of the function \(f(x) = x^{3} - 3x^{2}\) would tell us at which points the function is leveling off (slope of 0), or where it is increasing or decreasing.

Calculating a derivative is a fundamental process in calculus, and the power rule for differentiation is one of the essential tools for this purpose. By finding where the derivative equals zero, we can pinpoint where the function's rate of change switches from positive to negative or vice versa, which are known as critical points.

Power Rule for Differentiation

The power rule for differentiation is a quick method for finding derivatives of power functions. The general form states that if you have a function \(x^n\), where \(n\) is any real number, the derivative is \(nx^{n-1}\).

Looking at our exercise, when we differentiate \(x^{3}\) using the power rule, we bring down the exponent 3 to get \(3x^{3-1}\), which simplifies to \(3x^{2}\). Similarly, \(-3x^{2}\) becomes \(-6x^{2-1}\) or \(-6x\). The resulting derivative, \(f'(x) = 3x^{2} - 6x\), reveals how the function's slope changes at different values of \(x\). Understanding and applying the power rule is crucial for students tackling calculus problems efficiently.

Critical Points

Critical points of a function occur where the derivative is either zero or undefined. These points are essential because they often represent the peaks, valleys, or inflection points on a graph, marking where a function switches from increasing to decreasing or vice versa.

In our given function, the critical points are found by setting the derivative equal to zero. Thus \(3x^{2} - 6x = 0\), which can be factored to \(3x(x - 2) = 0\), giving us the critical points at \(x=0\) and \(x=2\). By testing intervals around these points, we can determine the function's behavior - it is increasing where the derivative is positive and decreasing where the derivative is negative. Recognizing critical points is a foundational skill in calculus for analyzing the behavior of functions.