Problem 18 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Graph the function given, labeling all \(x\) -intercepts, \(y\) intercepts, and the \(x\) - and \(y\) -coordinates of any local maximum and minimum points. $$ f(x)=x(x-2)^{2} $$

Step-by-Step Solution

Answer

The graph of the function \(f(x)=x(x-2)^{2}\) has x-intercepts at \(x = 0\) and \(x = 2\), a y-intercept at \(y = 0\), a local maximum at \(x = 0\), and a local minimum at \(x = 4/3\).

Step 1: Determine the x-intercepts

By setting \(f(x)=0\) and solving for \(x\), the x-intercepts can be determined. Simplifying \(f(x)=x(x-2)^{2}\) to \(x(x^2 - 4x + 4) = 0\), solutions for \(x\) can be found. These are \(x = 0\) and \(x = 2\).

Step 2: Determine the y-intercept

The y-intercept is determined where \(x = 0\). For \(f(x)=x(x-2)^{2}\), when \(x=0\), \(f(x) = 0(0-2)^{2} = 0\). So, the y-intercept is at \(y = 0\).

Step 3: Find the local maximum and minimum points

This is done by first finding the derivative of the function, setting it to 0 and solving for \(x\) to get the critical points. The derivative of \(f(x) = x(x-2)^{2}\) is \(f'(x) = 3x^2 - 4x\). Set \(f'(x) = 0\) to get: \(3x^2 - 4x = 0\), so \(x = 0\) or \(x = 4/3\).

Step 4: Identify local maximum and minimum points

To identify whether these points represent local maximums or minimums, the second derivative should be examined. The second derivative of \(f(x) = x(x-2)^{2}\) is \(f''(x) = 6x - 4\). Substituting \(x = 0\) into the second derivative, \(f''(0) = - 4\), which is less than 0, indicating that \(x = 0\) is a local maximum point. For \(x = 4/3\), \(f''(4/3) = 4\), which is greater than 0, indicating that \(x = 4/3\) is a local minimum point.

Key Concepts

Intercepts: Finding Where the Graph Crosses the Axes

Understanding intercepts is key when graphing functions. Intercepts are the points where the graph of a function crosses the axes.

• X-intercepts: These occur where the graph crosses the x-axis, meaning where the function equals zero. For our function, \(f(x)=x(x-2)^{2}\), we set it to zero: \(x(x-2)^{2}=0\). The solutions are \(x = 0\) and \(x = 2\), indicating x-intercepts at these points.


• Y-intercept: This is where the graph crosses the y-axis. It happens when \(x = 0\). Plugging 0 into our function gives \(f(0)=0(0-2)^{2}=0\). So, the y-intercept is at \(y = 0\).

Knowing intercepts helps you quickly sketch the basic shape of the graph by pinpointing where it interacts with the axes.

Local Extrema: Peaks and Valleys on the Graph

Local extrema refer to the highest and lowest points in a particular section of a graph; these are labeled as local maxima and minima.

• A local maximum is a point where the function changes from increasing to decreasing, resembling a peak on the graph.


• A local minimum is a point where the function shifts from decreasing to increasing, resembling a valley.

To identify these critical points for \(f(x) = x(x-2)^2\), we must find where the derivative equals zero. Solving the derivative \(f'(x) = 3x^2 - 4x = 0\) gives critical points at \(x = 0\) and \(x = \frac{4}{3}\). Determining whether these are maxima or minima involves additional steps.

Derivatives: Tools for Analyzing Function Behavior

Derivatives provide insights into the function's rate of change and are essential for identifying key features of the graph. For the function \(f(x)=x(x-2)^{2}\), the first derivative \(f'(x)=3x^2 - 4x\) assists in finding where the function slope is zero, highlighting potential local extrema.

• If \(f'(x) > 0\), the function is increasing.
• If \(f'(x) < 0\), the function is decreasing.

By setting \(f'(x) = 0\), we locate critical points at \(x = 0\) and \(x = \frac{4}{3}\). The nature of these critical points is further examined using the second derivative, which provides valuable insight into the concavity of the function at those points.

Second Derivative Test: Determining Concavity at Critical Points

The second derivative test helps in determining whether the critical points are local maxima or minima. It does so by evaluating the concavity of the function:

• If \(f''(x) > 0\), the graph is concave up, and the point is a local minimum.


• If \(f''(x) < 0\), the graph is concave down, and the point is a local maximum.

Calculating the second derivative \(f''(x) = 6x - 4\) helps clarify these points for \(f(x)=x(x-2)^2\). At \(x = 0\), \(f''(0) = -4\), indicating a local maximum. At \(x = \frac{4}{3}\), \(f''(\frac{4}{3}) = 4\), suggesting a local minimum. This test confirms the nature of the critical points, offering a clearer understanding of the graph's characteristics.