Question

For which of the following intervals does the function (in the image section) have a removable discontinuity?A. [−2.5,−1.5] B. [−1.5,−0.5] C. [−0.5,0.5] D. [0.5,1.5] E. [1.5,2.5]

Answer 1

Given the function below,

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

Let us now factorize the denominator

$x^4-2x^3-x^2+2x$

First of all, let us factorize x out from the denominator.

$\begin{gathered} x(\frac{x^4}{x}-\frac{2x^3}{x}-\frac{x^2}{x}+\frac{2x}{x}) \\ x(x^3-2x^2-x+2) \end{gathered}$

Therefore, x is a factor of the denominator.

Let us now factorize the remainder

$x^3-2x^2-x+2$

Let us substitute x = 1 into the function to confirm if it is a factor.

$\begin{gathered} x=1 \\ 1^3-2(1)^2-(1)+2 \\ 1-2-1+2=1+2-1-2=3-3=0 \end{gathered}$

Therefore, (x - 1) is a factor.

Let us now divide the function by (x - 1)

$\frac{x^3-2x^2-x+2}{x-1}=\frac{\left(x-2\right)\left(x+1\right)\left(x-1\right)}{x-1}=(x-2)(x+1)$

Hence, the factors of the denominator are,

$x(x-1)(x+1)(x-2)$

Therefore,

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

Now, looking at both the numerator and the denominator, we can observe that there is no common factor between the numerator and the denominator.

Hence, there is no removable discontinuity.