Problem 19 - 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. $$ f(x)=x^{2}(x-2) $$

Step-by-Step Solution

Answer

The function \(f(x) = x^{2}(x-2)\) has \(x\)-intercepts at \(0\) and \(2\), a \(y\)-intercept at \(0\), a local maximum point at \(0, 0\), and a local minimum point at \((4/3, -16/27)\).

Step 1: Find the x-intercepts

Set the function to zero and solve for \(x\), \(0 = x^{2}(x-2)\). This gives us two solutions, \(0\) and \(2\). These are the points where the function crosses the x-axis.

Step 2: Find the y-intercept

Substitute \(x = 0\) in the function \(f(x) = x^{2}(x-2)\). This gives us \(y=0\), meaning the function crosses the y-axis at \(0\).

Step 3: Find local maximum and minimum points

Take the derivative of the function \(f'(x) = 3x^{2}-4x\), and solve \(f'(x) = 0\), we get \(x=0\) and \(x=4/3\). To classify each point, take the second derivative \(f''(x) = 6x-4\). Substitute \(x=0\), \(f''(0) = -4 < 0\), so \(x = 0\) is a local maximum point. Substitute \(x = 4/3\), \(f''(4/3) = 4 > 0\), so \(x = 4/3\) is a local minimum point. Compute \(f(0) = 0\) and \(f(4/3) = -16/27\) to find the y-coordinates of the local maximum and minimum points respectively.

Step 4: Draw the graph

Plot the \(x\)-intercepts at \(0\) and \(2\), the \(y\)-intercept at \(0\), and the local maximum at \(0, 0\) and local minimum at \((4/3, -16/27)\) to sketch the graph of the function.

Key Concepts

x-intercepts

The x-intercepts of a function are the points where the graph crosses the x-axis. This happens when the value of the function is zero.
To find these points for the function \( f(x) = x^2(x-2) \), set it equal to zero:

• \( 0 = x^2(x - 2) \)

From this, we can factor out the terms to find the solutions. Setting each factor equal to zero, we get:

• \( x^2 = 0 \) which gives \( x = 0 \)
• \( x - 2 = 0 \) which gives \( x = 2 \)

So, the x-intercepts are at \(x = 0\) and \(x = 2\). These are fundamental for sketching the graph since they indicate where the curve meets the x-axis.

y-intercepts

The y-intercept is where the graph crosses the y-axis. This point is found by evaluating the function at \(x = 0\). For our example, substitute 0 into the function:\[f(0) = 0^2(0-2) = 0\]This tells us that the y-intercept is also at \(y = 0\).
The y-intercept often gives us the starting point for the graph, as it represents the value of the function when all other variables are zero.

local maximum and minimum points

Local maximum and minimum points represent the peaks and troughs in the graph. To find these points, we first need to determine the derivative of the function. For \( f(x) = x^2(x-2) \), the first derivative is:\[f'(x) = 3x^2 - 4x\]Setting this derivative to zero helps locate critical points:\[3x^2 - 4x = 0 \implies x(3x - 4) = 0\]

• \(x = 0\)
• \(x = \frac{4}{3}\)

These x-values are where the function has potential local maxima or minima. To determine which, we use the second derivative test. The y-values of the local maximum and minimum are found by substituting these x-values back into the function.

derivatives

Derivatives are a fundamental part of calculus used to find the rate at which functions change. They are helpful in identifying the slope of tangent lines, velocity of moving objects, and more.For the function \( f(x) = x^2(x-2) \), the derivative \( f'(x) = 3x^2 - 4x \) is calculated using the product rule and simplification. This derivative tells us the slope of the function at any given point on the graph, and it is essential for finding critical points where the function changes direction.

second derivatives

The second derivative is a further differentiation of the first derivative, providing information on the concavity of the function. It helps to classify local maximum and minimum points.For \( f(x) = x^2(x-2) \), the second derivative is:\[f''(x) = 6x - 4\]Use the second derivative to determine concavity:

• If \( f''(x) > 0 \), the graph is concave up at that point (indicating a local minimum).
• If \( f''(x) < 0 \), the graph is concave down at that point (indicating a local maximum).

For our specific points:

• \( f''(0) = -4 \) shows a local maximum at \(x = 0\).
• \( f''\left(\frac{4}{3}\right) = 4 \) shows a local minimum at \(x = \frac{4}{3}\).