Let [tex]$f(x)=\frac{4 x}{x^2+2 x-8}$[/tex].
| x | f(x) |
| :----- | :------- |
| 1.99 | ⋱ |
| 1.999 | C |
| 1.9999 | V |
| 2 | ⋱ |
| 2.0001 | D |
| 2.001 | D |
| 2.01 | D |
Answer (1)
Factor the denominator of f ( x ) as ( x + 4 ) ( x − 2 ) .
Evaluate f ( 1.999 ) to find C ≈ − 1332.8888 .
Evaluate f ( 1.9999 ) to find V ≈ − 13332.8889 .
Evaluate f ( 2.0001 ) to find D ≈ 13333.7778 .
Evaluate f ( 2.001 ) to find D ≈ 1333.7777 .
Evaluate f ( 2.01 ) to find D ≈ 133.7770 .
The function has a vertical asymptote at x = 2 .
As x approaches 2 from the left, f ( x ) approaches − ∞ , and as x approaches 2 from the right, f ( x ) approaches + ∞ .
Explanation
Problem Setup We are given the function f ( x ) = x 2 + 2 x − 8 4 x and a table of values for x near 2. We need to find the values of f ( x ) for x = 1.999 , 1.9999 , 2.0001 , 2.001 , 2.01 .
Factoring the Denominator First, let's factor the denominator of f ( x ) : x 2 + 2 x − 8 = ( x + 4 ) ( x − 2 ) So, we can rewrite the function as: f ( x ) = ( x + 4 ) ( x − 2 ) 4 x
Evaluating f(x) Now, we will evaluate f ( x ) at the given values of x :
For x = 1.999 :
C = f ( 1.999 ) = ( 1.999 + 4 ) ( 1.999 − 2 ) 4 ( 1.999 ) = ( 5.999 ) ( − 0.001 ) 7.996 ≈ − 1332.8888
For x = 1.9999 :
V = f ( 1.9999 ) = ( 1.9999 + 4 ) ( 1.9999 − 2 ) 4 ( 1.9999 ) = ( 5.9999 ) ( − 0.0001 ) 7.9996 ≈ − 13332.8889
For x = 2.0001 :
D 1 = f ( 2.0001 ) = ( 2.0001 + 4 ) ( 2.0001 − 2 ) 4 ( 2.0001 ) = ( 6.0001 ) ( 0.0001 ) 8.0004 ≈ 13333.7778
For x = 2.001 :
D 2 = f ( 2.001 ) = ( 2.001 + 4 ) ( 2.001 − 2 ) 4 ( 2.001 ) = ( 6.001 ) ( 0.001 ) 8.004 ≈ 1333.7777
For x = 2.01 :
D 3 = f ( 2.01 ) = ( 2.01 + 4 ) ( 2.01 − 2 ) 4 ( 2.01 ) = ( 6.01 ) ( 0.01 ) 8.04 ≈ 133.7770
Analyzing the Limit As x approaches 2 from the left, f ( x ) approaches − ∞ . As x approaches 2 from the right, f ( x ) approaches + ∞ . This indicates a vertical asymptote at x = 2 .
Final Values Therefore, the values are approximately: C ≈ − 1332.8888 V ≈ − 13332.8889 D 1 ≈ 13333.7778 (for x=2.0001) D 2 ≈ 1333.7777 (for x=2.001) D 3 ≈ 133.7770 (for x=2.01)
Examples
Understanding the behavior of functions near singularities, like the vertical asymptote in this problem, is crucial in fields like physics and engineering. For example, when analyzing the stress on a material near a sharp corner, the stress function might approach infinity, indicating a point of potential failure. Similarly, in electrical engineering, the electric field near a point charge can be modeled by a function with a singularity. Analyzing these functions helps engineers design safer and more reliable structures and devices.
