Problem 40 - College Algebra

Question

Refer to Table. $$ \begin{array}{|c|c|c|c|c|c|c|} \hline \boldsymbol{x} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \boldsymbol{f}(\boldsymbol{x}) & 5.1 & 6.3 & 7.3 & 7.7 & 8.1 & 8.6 \\ \hline \end{array} $$ Use the intersect feature to fi d the value of \(x\) for which \(f(x)=7\)

Step-by-Step Solution

Answer

The value of \(x\) when \(f(x) = 7\) is approximately 2.7.

Step 1: Understand the Table

The provided table gives a set of values for \(x\) and corresponding values of \(f(x)\). We are looking for an \(x\) where \(f(x) = 7\).

Step 2: Identify the Interval

Locate the values of \(f(x)\) that are closest to 7. From the table, note that \(f(3) = 7.3\) and \(f(2) = 6.3\). The value of \(7\) falls between these values.

Step 3: Use Linear Interpolation

Since \(f(x)\) seems to follow a nearly linear trend between consecutive points, use linear interpolation between \(f(2) = 6.3\) and \(f(3) = 7.3\) to find \(x\) corresponding to \(f(x) = 7\).

Step 4: Set up the Linear Interpolation Equation

The formula for linear interpolation is \(x = x_1 + \frac{y - y_1}{y_2 - y_1} (x_2 - x_1)\). Set \(x_1 = 2\), \(x_2 = 3\), \(y_1 = 6.3\), and \(y_2 = 7.3\). Substitute \(y = 7\) into the formula.

Step 5: Calculate the Interpolated Value

Substitute the values into the formula: \[ x = 2 + \frac{7 - 6.3}{7.3 - 6.3}(3 - 2) \]This simplifies to:\[ x = 2 + \frac{0.7}{1} \]Thus, \(x = 2 + 0.7 = 2.7\).

Step 6: Conclusion

After calculation, the value of \(x\) for which \(f(x) = 7\) is approximately \(x = 2.7\).

Key Concepts

Function Evaluation

Function evaluation is a fundamental concept in mathematics that entails substituting a value into a function to obtain a corresponding output. In simpler terms, it involves checking the behavior or outcome of a mathematical function when a certain input is given. This process is crucial because it helps in understanding how functions work.

• Consider a function represented as \( f(x) \).
• Evaluating it means substituting a specific \( x \) value to find \( f(x) \).
• In the example, determining \( f(x) = 7 \) when \( x = 2.7 \) involves function evaluation using linear interpolation.

By plugging in values into a function, you can make predictions or analyze different scenarios. In practical applications, especially in engineering or economics, function evaluation assists in modeling behaviors and examining systems.

Numerical Methods

Numerical methods are powerful techniques used to find approximate solutions to mathematical problems that might be too complex to solve analytically. Linear interpolation is a classic example of a numerical method.

• These methods transform complicated equations into simple computations.
• They are often employed in situations where a straightforward formulaic solution is not feasible.
• The linear interpolation method used in the exercise concludes that \( x \approx 2.7 \) when \( f(x) = 7 \).

Numerical methods play a pivotal role in scientific computing, helping us solve real-world problems by approximating solutions. They enable us to simulate phenomena like weather forecasting and even the behavior of financial markets.

Algebraic Equations

Algebraic equations form the backbone of mathematics, providing relationships between variables. Solving these equations means finding the values of the variables that make the equation true.

• The equation \( f(x) = 7 \) requires determining the value of \( x \).
• Linear interpolation creates an approximate algebraic equation for estimating unknown values between two known data points.
• In the given exercise, algebraic manipulation helped in forming and solving: \( x = 2 + \frac{0.7}{1}(3 - 2) \).

This example shows how algebraic equations, combined with numerical methods like interpolation, help solve practical problems where exact values might not be directly obtainable. Understanding these equations and how to manipulate them is key to various fields, from physics to finance.