Complete the table of values for the function [tex]f(x)=\frac{x-1}{x^2-1}[/tex]
| x | f(x) |
| :---- | :---- |
| 0.8 | |
| 0.9 | |
| 0.99 | |
| 0.999 | |
| 1 | |
| 1.001 | |
| 1.01 | |
| 1.1 | |
| 1.2 | |
Answer (1)
Simplify the function: f ( x ) = x 2 − 1 x − 1 = x + 1 1 for x = 1 .
Evaluate f ( x ) for each given x value using the simplified form.
At x = 1 , find the limit: lim x → 1 f ( x ) = 2 1 .
The completed table values are approximately: f ( 0.8 ) = 0.5556 , f ( 0.9 ) = 0.5263 , f ( 0.99 ) = 0.5025 , f ( 0.999 ) = 0.5003 , f ( 1 ) = 0.5 , f ( 1.001 ) = 0.4998 , f ( 1.01 ) = 0.4975 , f ( 1.1 ) = 0.4762 , f ( 1.2 ) = 0.4545 .
Explanation
Simplify the function We are given the function f ( x ) = x 2 − 1 x − 1 and asked to complete a table of values for various x values. First, we can simplify the function by factoring the denominator: x 2 − 1 = ( x − 1 ) ( x + 1 ) . Thus, f ( x ) = ( x − 1 ) ( x + 1 ) x − 1 . For x = 1 , we can cancel the ( x − 1 ) terms, so f ( x ) = x + 1 1 .
Evaluate the function Now, we can evaluate the simplified function f ( x ) = x + 1 1 at the given x values.
Calculate f(0.8) For x = 0.8 , f ( 0.8 ) = 0.8 + 1 1 = 1.8 1 = 0.5555555555555556 .
Calculate f(0.9) For x = 0.9 , f ( 0.9 ) = 0.9 + 1 1 = 1.9 1 = 0.5263157894736842 .
Calculate f(0.99) For x = 0.99 , f ( 0.99 ) = 0.99 + 1 1 = 1.99 1 = 0.5025125628140703 .
Calculate f(0.999) For x = 0.999 , f ( 0.999 ) = 0.999 + 1 1 = 1.999 1 = 0.5002501250625312 .
Calculate f(1) For x = 1 , the original function is undefined because we would have division by zero. However, we can find the limit as x approaches 1: lim x → 1 f ( x ) = lim x → 1 x + 1 1 = 1 + 1 1 = 2 1 = 0.5 .
Calculate f(1.001) For x = 1.001 , f ( 1.001 ) = 1.001 + 1 1 = 2.001 1 = 0.49975012493753124 .
Calculate f(1.01) For x = 1.01 , f ( 1.01 ) = 1.01 + 1 1 = 2.01 1 = 0.49751243781094534 .
Calculate f(1.1) For x = 1.1 , f ( 1.1 ) = 1.1 + 1 1 = 2.1 1 = 0.47619047619047616 .
Calculate f(1.2) For x = 1.2 , f ( 1.2 ) = 1.2 + 1 1 = 2.2 1 = 0.45454545454545453 .
Examples
Understanding functions and their values at specific points is crucial in many real-world applications. For example, in physics, you might use a function to model the trajectory of a projectile. By calculating the function's value at different times, you can determine the projectile's position at those times. Similarly, in economics, functions can model market trends, and evaluating the function at different points can help predict future market behavior. This exercise reinforces the fundamental skill of evaluating functions, which is essential for more advanced mathematical modeling and problem-solving.
