Problem 28 - Precalculus: Functions and Graphs

Question

Find all values of \(x\) such that \(f(x)>0\) and all \(x\) such that \(f(x)<0,\) and sketch the graph of \(f\) $$f(x)=x^{3}(x+1)^{2}(x-2)(x-4)$$

Step-by-Step Solution

Answer

\(f(x) > 0\) on \((-\infty, -1), (0, 2), (4, \infty)\); \(f(x) < 0\) on \((-1, 0), (2, 4)\).

Step 1: Identify the Roots

To find the intervals where the function changes signs, we first identify the roots of the function. The function is given as \(f(x) = x^3(x+1)^2(x-2)(x-4)\). Therefore, we set each factor equal to zero: \(x^3 = 0\), \((x+1)^2 = 0\), \(x-2 = 0\), \(x-4 = 0\). This gives us the roots as \(x = 0\), \(x = -1\), \(x = 2\), and \(x = 4\).However, note the multiplicity of the roots: \(x = 0\) has a multiplicity of 3, \(x = -1\) has a multiplicity of 2, \(x = 2\) has a multiplicity of 1, and \(x = 4\) has a multiplicity of 1.

Step 2: Determine the Sign Changes Around the Roots

We will now check the signs of \(f(x)\) in each interval defined by our roots. The critical intervals are \((-\infty, -1)\), \((-1, 0)\), \((0, 2)\), \((2, 4)\), and \((4, \infty)\).1. On \((-\infty, -1)\): Choose \(x = -2\). \(f(x) > 0\) because each factor will produce a positive or an even-power negative value which results in an overall positive product.2. On \((-1, 0)\): Choose \(x = -0.5\). Since \(x^3\) and \((x+1)^2\) remain positive while the other factors remain negative, \(f(x) < 0\).3. On \((0, 2)\): Choose \(x = 1\). Here, \(f(x) > 0\) as all factors remain positive.4. On \((2, 4)\): Choose \(x = 3\). The factor \((x-2)\) makes \(f(x) < 0\), causing negation.5. On \((4, \infty)\): Choose \(x = 5\). Each factor is positive, so \(f(x) > 0\).

Step 3: Sketch the Graph of the Function

To sketch the graph, start by plotting the zeros at \(x = -1, 0, 2,\) and \(4\) on a number line. Mark where the function changes its sign at these points.Consider the behavior around each root given their multiplicity:- At \(x = 0\), the function touches and goes through the axis (because of odd multiplicity).- At \(x = -1\), the function touches the axis but does not cross it (even multiplicity).- At \(x = 2\) and \(x = 4\), the function passes through the axis (both odd multiplicities).Between the intervals, sketch the function as changing the sign as calculated earlier, i.e., positive or above the x-axis for intervals \((-\infty, -1)\), \((0, 2)\), and \((4, \infty)\) and negative or below the x-axis for intervals \((-1, 0)\), and \((2, 4)\).

Step 4: Conclusion on Solutions

The function \(f(x) > 0\) on the intervals \((-\infty, -1)\), \((0, 2)\), and \((4, \infty)\), whereas \(f(x) < 0\) on the intervals \((-1, 0)\) and \((2, 4)\). These intervals represent all the regions where the function is strictly positive and strictly negative, respectively.

Key Concepts

Roots of Polynomials

When dealing with polynomials, one of the first steps is to find the roots, or zeros, of the polynomial function. Roots are the values of \(x\) for which the polynomial equals zero. In the polynomial \(f(x) = x^3(x+1)^2(x-2)(x-4)\), we have factors that can be set to zero to determine the roots:

• \(x^3 = 0\) gives \(x = 0\).
• \((x+1)^2 = 0\) gives \(x = -1\).
• \(x - 2 = 0\) gives \(x = 2\).
• \(x - 4 = 0\) gives \(x = 4\).

Finding roots helps us break the x-axis into different sections where the function's behavior can be analyzed. Knowing these roots, we can understand where the function changes direction or crosses the axis.

Multiplicity of Roots

Roots are not all equal in how they affect the graph of a polynomial. The multiplicity of a root tells us how often a particular root appears in the factorization of a polynomial and influences the graph's behavior at that point:

• A root with odd multiplicity indicates that the graph will cross the x-axis at that root. For example, \(x = 0\) with multiplicity 3 means the graph not only touches but also crosses the x-axis at this point.
• A root with even multiplicity, like \(x = -1\), means the graph will touch the x-axis but not cross it, instead it will "bounce" off.

For our function, \(x = 2\) and \(x = 4\) both have a multiplicity of 1, hence the graph sharply crosses these points. Understanding multiplicity is crucial for sketching accurate graphs and analyzing polynomial roots.

Graphical Representation of Functions

Visualizing a polynomial function helps interpret its behavior over its domain. The graph of a polynomial function like \(f(x) = x^3(x+1)^2(x-2)(x-4)\) can tell us a lot about where the function is positive or negative.First, plot the x-axis intersections using the roots. At each root, assess the behavior using multiplicities:

• At \(x = -1\), the function only touches the axis (even multiplicity) and does not cross.
• At \(x = 0\), it crosses the axis (odd multiplicity).

Between these plots, consider intervals: from \((-abla, -1)\), Plotting these provides a visual framework to observe patterns, especially where the graph rises above or falls below the x-axis. Using these signs at intervals, we sketch the polynomial function.

Sign Analysis of Intervals

Understanding where a polynomial function is positive or negative stems from sign analysis of intervals defined by its roots. By checking test values in each interval, you can determine whether the function outputs positive or negative values.The intervals divided by the roots of our function \(f(x)\) are:

• \((-abla, -1)\)
• \((-1, 0)\)
• \((0, 2)\)
• \((2, 4)\)
• \((4, abla)\)

For each interval, select a test value to calculate \(f(x)\):

• \((-abla, -1)\): Choose something like \(x = -2\), resulting in \(f(x) > 0\).
• \((-1, 0)\): Choose \(x = -0.5\), resulting in \(f(x) < 0\).
• \((0, 2)\): Choose \(x = 1\), resulting in \(f(x) > 0\).
• \((2, 4)\): Choose \(x = 3\), resulting in \(f(x) < 0\).
• \((4, abla)\): Choose \(x = 5\), resulting in \(f(x) > 0\).

Sign changes confirm understanding of where the function rises or falls, essential for comprehending polynomial inequalities.