Continuous Function II

Problem 145: At what point is the following function continuous?

(1) $\begin{equation*} y = \frac{x+1}{x^2-4x+3} \end{equation*}$

Solution: One can think of continuity as where is the function (1) no defined. That is, we need to find the domain of the function in order to come to this conclusion. That is,

(2) $\begin{equation*} y = \frac{x+1}{x^2-4x+3} = \frac{x+1}{(x-3)(x-1)}\ \quad \Rightarrow \quad x \ne 1, 3. \end{equation*}$

Note that we do not want $x$ to be equal to 1 or 3 since that would make our function (1) undefined (division by 0).

This means that the function (1) is continuous everywhere except at $x = 1$ and $x = 3$ .

Leave a Reply Cancel reply