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Question: If \[f\left( x \right) = \dfrac{{{x^3} - 4x}}{{{x^3} + {x^2} - 6x}}\], how do you find all the point...

If f(x)=x3−4xx3+x2−6xf\left( x \right) = \dfrac{{{x^3} - 4x}}{{{x^3} + {x^2} - 6x}}, how do you find all the points of discontinuity of f(x)f\left( x \right) ?

Explanation

Solution

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(x)=x3−4xx3+x2−6xf\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(x)=x(x−2)(x+2)x(x−2)(x+3)f\left( x \right) = \dfrac{{x\left( {x - 2} \right)\left( {x + 2} \right)}}{{x\left( {x - 2} \right)\left( {x + 3} \right)}}
f(x)=(x+2)(x+3)f\left( x \right) = \dfrac{{\left( {x + 2} \right)}}{{\left( {x + 3} \right)}}, with exclusions x≠0x \ne 0and x≠2x \ne 2.
There are removable discontinuities at x = 0 and x = 2 where both the numerator and the denominator are zero, so f(x)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.
lim⁡x→−3+f(x)=−∞\mathop {\lim }\limits_{x \to - {3^ + }} f\left( x \right) = - \infty
lim⁡x→−3−f(x)=+∞\mathop {\lim }\limits_{x \to - {3^ - }} f\left( x \right) = + \infty
We can also find in simplified manner as:
x3+x2−6x=0{x^3} + {x^2} - 6x = 0
Hence, finding the factors i.e.,
x(x2+x−6)=x(x−2)(x+3)=0x\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(x)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(x)f\left( x \right) is well defined and continuous.