Math Problem Statement
Find the value of y(2.9) from the following table, where the table provides the values of x: [0, 1, 2, 3, 4, 5] and corresponding values of y: [3, 3, 19, 69, 171, 343].
Solution
To find the value of from the table provided, we can use interpolation, which estimates the value of a function between known data points.
The table provides us with the following pairs of and :
\hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y & 3 & 3 & 19 & 69 & 171 & 343 \\ \hline \end{array}$$ Since $$2.9$$ lies between $$x = 2$$ and $$x = 3$$, we will use **linear interpolation** to estimate $$y(2.9)$$. The formula for linear interpolation between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is: $$y(x) = y_1 + \frac{(y_2 - y_1)}{(x_2 - x_1)} (x - x_1)$$ ### Applying the values: - $$x_1 = 2$$, $$y_1 = 19$$ - $$x_2 = 3$$, $$y_2 = 69$$ - $$x = 2.9$$ Using the formula: $$y(2.9) = 19 + \frac{(69 - 19)}{(3 - 2)}(2.9 - 2)$$ $$y(2.9) = 19 + 50 \times 0.9$$ $$y(2.9) = 19 + 45 = 64$$ ### Therefore, $$y(2.9) \approx 64$$. Would you like more details on interpolation, or have any questions? Here are 5 related questions to deepen your understanding: 1. What is the general difference between linear and polynomial interpolation? 2. How would the result change if quadratic interpolation was used instead of linear? 3. Can you perform interpolation for values outside the given range (e.g., extrapolation)? 4. How is interpolation used in real-world applications, such as data analysis or computer graphics? 5. What are the limitations of using interpolation with non-linear data? **Tip**: Always check the interval size between known data points—if they change significantly, linear interpolation might not be the best approximation.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Interpolation
Linear Interpolation
Formulas
Linear interpolation formula: y(x) = y_1 + ((y_2 - y_1) / (x_2 - x_1)) * (x - x_1)
Theorems
Linear Interpolation Theorem
Suitable Grade Level
Grades 9-12