Problem 29 - Precalculus : Building Concepts and Connections

Question

Let \(f(x)=\frac{2 x^{2}-1}{x^{2}}\) (a) Fill in the following table for values of \(x\) near zero. What do you observe about the value of \(f(x)\) as \(x\) approaches zero from the right? from the left?$$\begin{array}{|c|c|c|c|c|c|} \hline x & -0.5 & -0.1 & -0.01 & 0.01 & 0.1 \\ \hline f(x) & & & & & \end{array}$$(b) Complete the following table. What happens to the value of \(f(x)\) as \(x\) gets very large and positive? $$\begin{array}{ccccc} x & 10 & 50 & 100 & 1000 \\ \hline f(x) & & & & \end{array}$$ (c) Complete the following table. What happens to the value of \(f(x)\) as \(x\) gets very large and negative? (TABLE CAN'T COPY)

Step-by-Step Solution

Answer

As \(x\) approaches zero from the right, \(f(x)\) goes to positive infinity. As \(x\) approaches zero from the left, \(f(x)\) goes to negative infinity. As \(x\) get very large and positive or very large and negative, \(f(x)\) tends to 2.

Step 1: Find Function Values

Substitute the given \(x\)-values into the function, \(f(x)=\frac{2 x^{2}-1}{x^{2}}\). Calculate \(f(x)\) for \(x\) equals -0.5, -0.1, -0.01, 0.01 and 0.1. Carefully compute each and record the values in the table.

Step 2: Interpret Results

Analyze the results. As \(x\) approaches zero from the right, \(f(x)\) tends to positive infinity. As \(x\) approaches zero from the left, \(f(x)\) tends to negative infinity.

Step 3: Find Function Values for Positive X

Again, substitute \(x\)-values into the function, but this time use large positive numbers, namely: 10, 50, 100 and 1000. Compute each value.

Step 4: Interpret Second Table Results

Examine these results. As \(x\) gets very large and positive, \(f(x)\) tends to 2.

Step 5: Find Function Values for Negative X

Substitute large negative \(x\)-values into the function. Compute the values.

Step 6: Interpret Third Table Results

Look at these results. As \(x\) gets very large and negative, \(f(x)\) also tends to 2.

Key Concepts

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as the input values approach a certain point or move towards infinity and beyond. For the function \( f(x) = \frac{2x^2 - 1}{x^2} \), understanding its asymptotic behavior helps in grasping what happens when \( x \) approaches extreme values such as zero, or becomes very large, either positively or negatively.

• As \( x \to 0^+ \) (approaches zero from the right), the numerator \( 2x^2 - 1 \) remains negative, while the denominator \( x^2 \) approaches zero, resulting in the function value tending towards positive infinity.
• Conversely, as \( x \to 0^- \) (zero from the left), the function value approaches negative infinity. This reveals the vertical asymptote at \( x = 0 \), where the function grows unbounded.
• As \( x \to \pm \infty \), both the top and bottom of the fraction are dominated by \( 2x^2 \). This means the function simplifies to approximately \( 2 \), indicating a horizontal asymptote at \( y = 2 \).

Understanding asymptotic behavior aids in predicting function behavior without plotting every single point.

Function Analysis

Function analysis involves breaking down a function to understand its different characteristics such as domain, range, intercepts, and continuity. In the case of \( f(x) = \frac{2x^2 - 1}{x^2} \), several aspects should be considered:

• Domain: The domain excludes \( x = 0 \) because division by zero is undefined. So, \( x eq 0 \).
• Range: As discussed in the asymptotic behavior, the range of \( f(x) \) is all real numbers except possibly close to a discontinuity.
• Intercepts: Calculating intercepts shows that there are no x-intercepts because \( 2x^2 - 1 eq 0 \) for real \( x \).

The analysis tells us the behavior of the function as explored through different situations (positive, negative, origin), thus offering a complete view of its character and graph.

Precalculus Concepts

Precalculus lays the foundation for understanding calculus by dealing with basic elements of functions, limits, and continuity. Applying these precalculus concepts helps in analyzing functions like \( f(x) = \frac{2x^2 - 1}{x^2} \).

• Limits: Limits form the cornerstone of calculus and seek to understand a function’s behavior close to a particular point. When investigating what happens as \( x \to 0 \) or \( x \to \infty \), recognizing the limit can clarify that \( f(x) \to \pm \infty \) near zero and \( f(x) \to 2 \) far off along the x-axis.
• Simplification of Expressions: Recognizes that symmetric properties of a function allow it to be approximated under certain conditions for easy computation, such as observing that \( f(x) \approx 2 \) for large \( |x| \).
• Continuity and Discontinuity: Understanding that \( f(x) \) is discontinuous at \( x=0 \) gives insights into its behavior around this point and stylistic jumps or breaks at where it is not defined.

Grasping precalculus concepts solidifies a student’s comprehension of not just the specific problem at hand but also broadens the ability to tackle more complex functions and calculus-centered tasks.