Question

For which of the following intervals does the function (insert image) 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

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

First, we need to factorize the quadratic expressions. To do this we need to find the roots of each quadratic expression. In the function of the numerator, the coefficients are: a = 1, b = -4, and c = 3. Applying the quadratic formula, the roots are:

$\begin{gathered} x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_{1,2}=\frac{4\pm\sqrt[]{(-4)^2-4\cdot1\cdot3}}{2\cdot1} \\ x_{1,2}=\frac{4\pm\sqrt[]{4}}{2} \\ x_1=\frac{4+2}{2}=3 \\ x_2=\frac{4-2}{2}=1 \end{gathered}$

Then, we can express the function with its roots as follows:

$\begin{gathered} ax^2+bx+c=a(x-x_1)(x-x_2) \\ x^2-4x+3=(x-3)(x-1) \end{gathered}$

In the first function of the denominator, the coefficients are: a = 1, b = 1, and c = -2. Applying the quadratic formula, the roots are:

$\begin{gathered} x_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ x_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ x_1=\frac{-1+3}{2}=1 \\ x_2=\frac{-1-3}{2}=-2 \end{gathered}$

Then, we can express the function with its roots as follows:

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

In the second function of the denominator, the coefficients are: a = 1, b = -2, and c = -3. Applying the quadratic formula, the roots are:

$\begin{gathered} x_{1,2}=\frac{2\pm\sqrt[]{(-2)^2-4\cdot1\cdot(-3)}}{2\cdot1} \\ x_{1,2}=\frac{2\pm\sqrt[]{16}}{2} \\ x_1=\frac{2+4}{2}=3 \\ x_2=\frac{2-4}{2}=-1 \end{gathered}$

Then, we can express the function with its roots as follows:

$x^2-2x-3=(x-3)(x+1)$

Substituting the equivalent expression into the original rational function, we get:

$\begin{gathered} f(x)=\frac{(x-3)(x-1)}{(x-1)(x+2)(x-3)(x+1)} \\ \text{ Simplifying:} \\ f(x)=\frac{1}{(x+2)(x+1)} \end{gathered}$

After the simplification, the discontinuities at x = 3 and x = 1 have been removed. x = 1 belongs to the interval [0.5, 1.5]