Problem 30 - College Algebra with Corequisite Support

Question

For the following exercises, refer to \(\underline{\text { Table }} 7 .\) $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \boldsymbol{x} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \boldsymbol{f}(\boldsymbol{x}) & 1125 & 1495 & 2310 & 3294 & 4650 & 6361 \\ \hline \end{array} $$ Use the intersect feature to find the value of \(x\) for which \(f(x)=4000\)

Step-by-Step Solution

Answer

The value of \(x\) for which \(f(x)=4000\) is approximately 4.52.

Step 1: Understand the Problem

We have a table with values of \(x\) and corresponding \(f(x)\) values. We need to find the value of \(x\) for which \(f(x) = 4000\).

Step 2: Analyze the Table

The given table contains values of \(f(x)\) that increase with \(x\). \(f(x)\) rises from 1125 at \(x = 1\) to 6361 at \(x = 6\). Since 4000 is within this range, we'll need to estimate \(x\) based on the closest \(f(x)\) values in the table.

Step 3: Locate Closest Values

In the table, \(f(x)=3294\) for \(x=4\) and \(f(x)=4650\) for \(x=5\). Since 4000 is between 3294 and 4650, the value of \(x\) that results in \(f(x)=4000\) will be between 4 and 5.

Step 4: Use Interpolation (if necessary)

We use linear interpolation between \(x=4\) and \(x=5\) to estimate the \(x\) for \(f(x)=4000\). Linear interpolation formula is:\[ x = x_1 + \frac{y - y_1}{y_2 - y_1} \cdot (x_2 - x_1) \]Substitute \(x_1 = 4\), \(x_2 = 5\), \(y_1 = 3294\), \(y_2 = 4650\), \(y = 4000\):\[ x = 4 + \frac{4000 - 3294}{4650 - 3294} \cdot (5 - 4) \]\[ x \approx 4 + \frac{706}{1356} \]\[ x \approx 4.52 \]

Step 5: Verify Results

Re-check the interpolation calculation. The value \(x \approx 4.52\) seems reasonable as it fits between the given values.

Key Concepts

Linear Interpolation

Linear interpolation is a popular method used to estimate unknown values between two known values. It works by connecting these two values with a straight line and assuming that this line accurately represents the data in that interval.
Imagine you have two points on a graph, \((x_1, y_1)\) and \((x_2, y_2)\). Linear interpolation allows you to find an intermediate point, \(x\), for which you want to estimate its corresponding value, \(y\).

• It is useful when data is linear or can be approximated as linear over small ranges.
• The formula for linear interpolation is \[ y = y_1 + \frac{x - x_1}{x_2 - x_1} \cdot (y_2 - y_1) \]

In our example, we applied linear interpolation to find the \(x\) for which \(f(x) = 4000\), estimating \(x\) to be approximately 4.52. This estimate is precise within our data points \(x = 4\) and \(x = 5\), especially since the data changes smoothly between them.

Piecewise Functions

Piecewise functions are functions that use different expressions for different intervals of the input variable. These functions allow us to model situations where data behaves differently in separate sections.
For example, the table provided gives values of \(f(x)\) at specific integer intervals of \(x\). Between these intervals, the function that might describe \(f(x)\) could change behavior or slope.

• Piecewise functions are defined using different formulas for different portions of their domain.
• The function can be linear in some sections and non-linear in others.

In the problem's context, linear interpolation was used as a simple piecewise approach where each section between known data points was assumed to behave linearly. Here, the piecewise function is linear between each integer \(x\). This approach simplifies the process of handling complex functions over smaller expected ranges.

Estimating Values

Estimating values involves choosing a method to infer unknown quantities based on known data. It is a crucial skill wherever exact measurements are unavailable or difficult to obtain.
Many scenarios involve estimating values, such as using interpolation, extrapolation, or averaging nearby data points.

• Interpolation, specifically linear interpolation as shown, can significantly reduce errors in estimation within observed data ranges.
• Ensure that the choice of estimation method suits the data characteristics (e.g., smoothness, distribution).

In this exercise, estimating the value of \(x\) when \(f(x) = 4000\) helped us find a non-tabulated value by balancing the values on either side of 4000 using interpolation. Thus, interpolation provided a valuable estimate that is often sufficient for practical purposes.