Problem 465 - Precalculus

Question

For the following exercises, refer to Table 4.28. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {1} & {2} & {3} & {4} & {5} & {6} \\\ \hline f(x) & {5.1} & {6.3} & {7.3} & {7.7} & {8.1} & {8.6} \\\ \hline\end{array}$$ Use the intersect feature to find the value of \(x\) for which \(f(x)=7\).

Step-by-Step Solution

Answer

The value of \(x\) is approximately 3.25 where \(f(x) = 7\).

Step 1: Understand the Problem

We need to find the value of \(x\) for which \(f(x) = 7\). We will use the intersect feature, which involves looking at the given function values in the table and identifying the \(x\) value where \(f(x)\) is equal to 7.

Step 2: Locate Relevant Values

Examine the function values \(f(x)\) in the table. We see the relevant values are around \(7.3\) for \(x=3\) and \(7.7\) for \(x=4\). The value \(7\) lies between these two function values.

Step 3: Find Intersection

Since the table shows that \(f(x)\) transitions from \(7.3\) to \(7.7\), the function \(f(x) = 7\) is crossed between \(x=3\) and \(x=4\). We can interpolate to find a more exact value within this range.

Step 4: Linear Interpolation

Using linear interpolation to estimate where \(f(x) = 7\), compute the proportional distance between \(7.3\) and \(7.7\). The formula is given by: \(x = 3 + \frac{7 - 7.3}{7.7 - 7.3} \times (4 - 3) \). Calculate the expression to find \(x\).

Step 5: Calculate the Interpolated Value

Compute the interpolated value: \(x = 3 + \frac{-0.3}{0.4} = 3 + (-0.75) = 3.25\). Therefore, \(f(x) = 7\) when \(x \approx 3.25\).

Key Concepts

Function Values

Function values are vital in understanding how a function behaves at specific points. The function given in the exercise is denoted by \( f(x) \) and its purpose is to assign a corresponding output for each input \( x \). In the table provided, these values are listed for integers from 1 to 6. The output values are \( 5.1, 6.3, 7.3, 7.7, 8.1, \) and \( 8.6 \).
Function values allow us to summarize data efficiently and view the trend of the function as \( x \) changes. They play a crucial role in interpolation because they provide the known points we use to estimate missing values in between.
When we have a target output, such as \( f(x) = 7 \), checking function values helps us find relevant points in the table, between which interpolation can occur.

Interpolation

Interpolation is a mathematical process used to estimate unknown values that fall within the range of known data points. In linear interpolation, we assume that the function between two points is a straight line.
This technique is employed in our problem to find an \( x \) value for which \( f(x) = 7 \). Since 7 is not directly listed in the table, interpolation helps us estimate the \( x \) that would correspond to \( f(x) = 7 \), given that \( f(x) \) transitions from \( 7.3 \) at \( x=3 \) to \( 7.7 \) at \( x=4 \).
We use the formula: \[x = x_1 + \frac{(y-y_1)}{(y_2-y_1)} \times (x_2-x_1) \]Here, \( y \) is the desired function value, whereas \( y_1 \) and \( y_2 \) are the known function values. This method makes linear interpolation straightforward and precise.

X and Y Coordinates

The concepts of \( x \) and \( y \) coordinates are foundational in understanding how data points are represented graphically. In tables, the \( x \) coordinates represent the input values, while the \( y \) coordinates represent the corresponding function values \( f(x) \).
In our problem, the table distinctly presents \( x \) values ranging from 1 to 6, while the \( y \) values \( f(x) \) are \( 5.1, 6.3, 7.3, 7.7, 8.1, \) and \( 8.6 \).
When using interpolation, understanding the positioning of these coordinates is crucial. For the task at hand, identifying the correct \( x \, (x_1 = 3, x_2 = 4) \) and \( y \, (y_1 = 7.3, y_2 = 7.7) \) values from the table helps set boundaries within which interpolation can be performed.
Recognizing this relationship aids in plotting these points on a graph, should a visual representation be needed, providing a clearer picture of how the function behaves between points.