Problem 52 - College Algebra and Calculus: An Applied Approach

Question

Use a graphing utility to graph \(f\) and \(f^{\prime}\) on the interval \([-2,2]\). $$ f(x)=x^{2}(x+1) $$

Step-by-Step Solution

Answer

The graph of the function \(f(x)\) appears as a curve with a hump that increases steadily. The graph of \(f^{\prime}(x)\) starts from a negative value at -2 and increases until reaching 0 at -1. It again starts decreasing and reaches 0 at 0, and then starts to increase steadily. The analysis shows how the derivative provides information about the slope of the original function.

Step 1: Simplify the Function

First, multiplication and distribution operations must be performed to simplify the function \(f(x) = x^{2}(x+1)\) given. After distribution, the function simplifies to \(f(x) = x^{3} + x^{2}\).

Step 2: Find the Derivative

Next, the derivative of the function \(f(x)\) is calculated by taking the derivative of each term with respect to \(x\). The derivative of \(x^{3}\) is \(3x^{2}\) and the derivative of \(x^{2}\) is \(2x\). So, the derivative of \(f(x)\), \(f’(x)\), is \(3x^{2} + 2x\).

Step 3: Graph the Functions

The functions \(f(x)\) and \(f^{\prime}(x)\) should then be graphed on the interval \([-2,2]\) using a suitable graphing tool. Plot the graphs of \(f(x) = x^{3} + x^{2}\) and \(f’(x) = 3x^{2} + 2x\).

Key Concepts

Derivative Calculation

The process of derivative calculation involves finding the rate at which a function changes at any point on its curve. It's a fundamental concept in calculus, often noted as the function's slope at a given point. When it comes to polynomial functions such as f(x) = x^3 + x^2, calculating the derivative is a straightforward application of known rules.

For example, the power rule is apply for the derivative of any term like x^n, which states that the derivative is n*x^(n-1). Applying this to our given function term-by-term, we obtain f'(x) = 3x^2 + 2x. This gives us the velocity of f(x) for any x, indicating how steep or flat the function's graph is at that point. Understanding this calculation is crucial because it allows students to anticipate the behavior of the function without even graphing it, telling us where the function increases, decreases, or has any turning points.

Function Simplification

Function simplification is the process of breaking down a complex expression into the simplest form possible, often making it easier to work with, particularly for differentiation or integration. Simplifying functions may involve expanding multiplications, factoring, combining like terms, or using algebraic identities. For our given function, f(x) = x^2(x + 1), simplification is achieved by multiplying the polynomial x^2 across the binomial (x + 1), resulting in f(x) = x^3 + x^2.

This simplified form is significantly more manageable when calculating the derivative because it breaks down the function into monomials, each of which can be tackled separately using the power rule. Simplification before taking a derivative is a valuable step because it reveals the fundamental components of the function, which can influence its graph and the properties of its derivative.

Graphing Utility

A graphing utility is an essential tool for providing visual representations of functions and their derivatives. It bridges the gap between the abstract calculations and concrete comprehension of the function's behavior. With the functions f(x) = x^3 + x^2 and its derivative f'(x) = 3x^2 + 2x, a graphing utility can help illustrate how the derivative reflects the slope of the original function at any given point.

When students graph these functions, they should look for crucial points such as where the derivative crosses the x-axis, which correspond to the peaks and valleys of the original function. It's also important to observe the intervals where the derivative is positive (where f(x) is increasing) and where it's negative (where f(x) is decreasing). The visual output from a graphing utility can make the abstract concept of derivatives far more tangible and aid in deeper understanding.