Question

Which statement is true about the discontinuities of the function f(x)? f (x) = StartFraction x squared minus 4 Over x cubed minus x squared minus 2 x EndFraction There is a hole at x = 2. There are asymptotes at x = 0 and x = –1. There are asymptotes at x = 0 and x = –1 and a hole at (2, two-thirds). There are holes at x = 0 and x = –1 and an asymptote at x = 2.

Answer 1

The discontinuities of the function is at the value of x equal to 0 and -1. The correct statement is there are asymptotes at x = 0 and x = –1.

What is the discontinuities of the function?

A function is discontinues at the point which is outside of the domain of the function. At this point in a graph , there are asymptotes.

The given function is,

$f(x)=\dfrac{x^2-4}{x^3-x^2-2x}$

Factorize the numerator and the denominator to solve it further.

$f(x)=\dfrac{x^2-2^2}{x(x^2-x-2)}\\f(x)=\dfrac{(x-2)(x+2)}{x(x+1)(x-2)}\\f(x)=\dfrac{(x+2)}{x(x+1)}$

Equate the factors of denominator equal to zero, to find the discontinuity.

$x(x+1)=0\\x=0,x=-1$

Hence, the discontinuities of the function is at the value of x equal to 0 and -1. The correct statement is there are asymptotes at x = 0 and x = –1.

Learn more about the discontinuities of the function here;

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