Question

If \[f\left( x \right) = \dfrac{{{x^3} - 4x}}{{{x^3} + {x^2} - 6x}}\], how do you find all the points of discontinuity of \[f\left( x \right)\] ?

Answer 1

Hint:As the given function is a rational function, to find out the discontinuity of the function, start by factoring the numerator and denominator of the function. A point of discontinuity occurs when a number is both a zero of the numerator and denominator and a rational function is continuous on its domain. The points of discontinuity for a rational function are the points outside the domain. Hence, by this we can find the points of discontinuity.

Complete step by step answer:
Let us write the given function:
\[f\left( x \right) = \dfrac{{{x^3} - 4x}}{{{x^3} + {x^2} - 6x}}\]
From the given function factor out the numerator and denominator to identify their zeros to determine the points of discontinuity and their types.
\[f\left( x \right) = \dfrac{{x\left( {x - 2} \right)\left( {x + 2} \right)}}{{x\left( {x - 2} \right)\left( {x + 3} \right)}}\]
\[f\left( x \right) = \dfrac{{\left( {x + 2} \right)}}{{\left( {x + 3} \right)}}\], with exclusions \[x \ne 0\]and \[x \ne 2\].
There are removable discontinuities at x = 0 and x = 2 where both the numerator and the denominator are zero, so \[f\left( x \right)\]is undefined at those points, but the left and right limits agree. There is a simple pole at x = -3, where the denominator is zero and the numerator is non-zero.
\[\mathop {\lim }\limits_{x \to - {3^ + }} f\left( x \right) = - \infty \]
\[\mathop {\lim }\limits_{x \to - {3^ - }} f\left( x \right) = + \infty \]
We can also find in simplified manner as:
\[{x^3} + {x^2} - 6x = 0\]
Hence, finding the factors i.e.,
\[x\left( {{x^2} + x - 6} \right) = x\left( {x - 2} \right)\left( {x + 3} \right) = 0\]
The points outside the domain are: 0, 2 and -3.

Therefore, the function \[f\left( x \right)\] has discontinuities at 0, 2 and -3.

Note:The key point to solve the given function is, as it is a rational function, we can see that, the only infinite limit occurs at x = −3. There is a non-removable (infinite) discontinuity at −3. Therefore, the discontinuities at 0 and 2 are removable. A function being continuous at a point means that the two-sided limit at that point exists and is equal to the function's value, discontinuity is when the two-sided limit exists, but isn't equal to the function's value. Apart from these discontinuities, \[f\left( x \right)\] is well defined and continuous.