Problem 74 - College Algebra with Corequisite Support

Question

For the following exercises, use a calculator to graph \(f(x)\). Use the graph to solve \(f(x)>0\). $$ f(x)=\frac{(x+3)^{2}}{(x-1)^{2}(x+1)} $$

Step-by-Step Solution

Answer

\(f(x) > 0\) for \(x \in (-3, -1) \cup (1, \infty)\).

Step 1: Understand the Function

The function given is \( f(x) = \frac{(x+3)^2}{(x-1)^2(x+1)} \). It is a rational function, meaning it is the ratio of two polynomials. The numerator is \((x+3)^2\), and the denominator is \((x-1)^2(x+1)\).

Step 2: Identify Critical Points

To determine where the function is greater than zero, first find the critical points, which include the zeros of the numerator and the undefined points from the denominator. The zeros are where \((x+3)^2 = 0\), which simplifies to \(x = -3\). The function is undefined at \(x = 1\) and \(x = -1\).

Step 3: Analyze the Behavior Around Critical Points

Evaluate the sign of \(f(x)\) in each interval determined by the critical points: \((-\infty, -3)\), \((-3, -1)\), \((-1, 1)\), and \((1, \infty)\). Check the behavior of the function in each interval using test points.

Step 4: Use a Calculator to Graph the Function

With a graphing calculator, plot \(f(x)\) to visually confirm where the graph is above the x-axis (i.e., where the function is positive). Observe these intervals on the graph.

Step 5: Determine Intervals where \(f(x) > 0\)

The graph shows that \(f(x) > 0\) in the interval \((-3, -1)\) and \((1, \infty)\) because the graph is above the x-axis in these regions.

Key Concepts

Critical Points

Critical points in a function are the points where the function changes its behavior. For rational functions like our function \(f(x) = \frac{(x+3)^2}{(x-1)^2(x+1)}\), critical points can include zeros of the numerator and points where the function is undefined (which occur at the zeros of the denominator).
- **Zeros of the Numerator**: To find where the numerator is zero, solve \((x + 3)^2 = 0\). This occurs at \(x = -3\).- **Undefined Points from the Denominator**: The function becomes undefined where the denominator is zero, \((x - 1)^2(x + 1) = 0\), which simplifies to \(x = 1\) and \(x = -1\).
Identifying these critical points is critical because they help divide the number line into intervals where the behavior of the function can be analyzed.

Zeros of Functions

The zeros of a function are the x-values where the function equals zero. For rational functions, these zeros are determined by the numerator. Here, our function \(f(x) = \frac{(x+3)^2}{(x-1)^2(x+1)}\) has the zero at \(x = -3\). This point signifies that the graph intersects the x-axis at that location. Identifying such zeros is the first step in solving inequalities like \(f(x) > 0\), as we need to test the sign of the function around these points.

Interval Analysis

Interval analysis involves examining sections of the x-axis divided by critical points. For our function, we identified the intervals:

• \((-\infty, -3)\)
• \((-3, -1)\)
• \((-1, 1)\)
• \((1, \infty)\)

We test values within each interval to determine if the function is positive or negative. For instance, in the interval \((-3, -1)\), pick any x-value, such as \(-2\), to substitute into \(f(x)\). This helps ascertain that \(f(x)\) is positive in this region.
This detailed interpretation is essential for understanding where \(f(x) > 0\) or \(f(x) < 0\) across different segments of the x-axis.

Graphical Solutions

Graphical solutions involve using visual graphs to understand the behavior of functions. By plotting \(f(x) = \frac{(x+3)^2}{(x-1)^2(x+1)}\) on a graphing calculator, you can visually identify where the function is above the x-axis, indicating \(f(x) > 0\). This graphical method provides an intuitive way to confirm our calculations.
Look at the graph to see:

• Where the graph crosses the x-axis, which corresponds to zeros of the function at \(x = -3\).
• The x-intervals where the graph is positioned above the x-axis, specifically \((-3, -1)\) and \((1, \infty)\) where \(f(x)\) is positive.

Graphing allows for quick verification of the intervals found through analytical methods and can reveal additional insights such as asymptotic behavior due to the undefined points \(x = 1\) and \(x = -1\).