Which statement is true about the discontinuities of the function [tex]f(x)[/tex]?
[tex]f(x)=\frac{x^2-4}{x^3-x^2-2 x}[/tex]
A. There is a hole at [tex]x=2[/tex].
B. There are asymptotes at [tex]x=0[/tex] and [tex]x=-1[/tex].
C. There are asymptotes at [tex]x=0[/tex] and [tex]x=-1[/tex] and a hole at [tex](2, \frac{2}{3})[/tex].
D. There are holes at [tex]x=0[/tex] and [tex]x=-1[/tex] and an asymptote at [tex]x=2[/tex].
Answer (2)
The function f ( x ) = x 3 − x 2 − 2 x x 2 − 4 has vertical asymptotes at x = 0 and x = − 1 , and a hole at the point ( 2 , 3 2 ) . Therefore, the correct option is C. There are asymptotes at x = 0 and x = − 1 and a hole at ( 2 , 3 2 ) .
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Factor the numerator and denominator of the function: f ( x ) = x 3 − x 2 − 2 x x 2 − 4 = x ( x − 2 ) ( x + 1 ) ( x − 2 ) ( x + 2 ) .
Simplify the function by canceling the common factor ( x − 2 ) : f ( x ) = x ( x + 1 ) x + 2 for x = 2 .
Identify the vertical asymptotes at x = 0 and x = − 1 from the simplified function.
Determine the hole at x = 2 , and find its y -coordinate by plugging x = 2 into the simplified function: f ( 2 ) = 2 ( 2 + 1 ) 2 + 2 = 3 2 . Thus, the hole is at ( 2 , 3 2 ) . The correct statement is: There are asymptotes at x = 0 and x = − 1 and a hole at ( 2 , 3 2 ) .
Explanation
Problem Analysis We are given the function f ( x ) = x 3 − x 2 − 2 x x 2 − 4 and asked to determine which statement is true about its discontinuities.
Factoring First, we factor the numerator and denominator to simplify the function:
Numerator: x 2 − 4 = ( x − 2 ) ( x + 2 )
Denominator: x 3 − x 2 − 2 x = x ( x 2 − x − 2 ) = x ( x − 2 ) ( x + 1 )
So, f ( x ) = x ( x − 2 ) ( x + 1 ) ( x − 2 ) ( x + 2 )
Simplifying We can simplify the function by canceling the common factor ( x − 2 ) :
f ( x ) = x ( x + 1 ) x + 2 , for x = 2
Identifying Asymptotes and Holes The original function is undefined when x = 0 , x = − 1 , and x = 2 . After simplification, the function is undefined when x = 0 and x = − 1 . This means there are vertical asymptotes at x = 0 and x = − 1 . The factor ( x − 2 ) was canceled, which means there is a hole at x = 2 .
Finding the Hole's Coordinates To find the y -coordinate of the hole, we plug x = 2 into the simplified function:
f ( 2 ) = 2 ( 2 + 1 ) 2 + 2 = 2 ( 3 ) 4 = 6 4 = 3 2
So, there is a hole at ( 2 , 3 2 ) .
Conclusion Therefore, there are asymptotes at x = 0 and x = − 1 and a hole at ( 2 , 3 2 ) .
Examples
Understanding discontinuities in functions is crucial in fields like physics and engineering. For example, when modeling the behavior of electrical circuits, discontinuities can represent sudden changes in voltage or current. Identifying and analyzing these points helps engineers design robust and reliable systems. Similarly, in fluid dynamics, discontinuities can model shock waves or abrupt changes in fluid density, which are critical for designing efficient aircraft and pipelines.
