Finding discontinuity in a function with constants.

Question

Question - need to find discontinuity points for the following function for all $\alpha,\beta\in\mathbb{R}$ $$f(x)=\begin{cases} \dfrac{x^{2}+\alpha x+\beta}{x^{2}-8x+15} && x\neq3,5\\[1ex] 0 && x=3,5 \end{cases}$$

I am having trouble arranging myself as I dont know if I should split this into few cases ? if yes what exact cases ? I know to prove all of the limits that can appear here but these constants caught me off guard.

I did the first case is when $ \alpha=\beta=0 $

But from there on - no idea. Will appreciate some help.

Answer 1

The possible points of discontinuity are $3$ and $5$. Now, for which $\alpha$ and $\beta$ do you have discontinuity at these points?

So notice that:

• $f$ is continuous at $3$ iff $\lim_{x\to 3} f(x)=0$, iff $x^2+\alpha x+\beta =(x-3)^2$, cause in such a case \begin{align*} \lim_{x\to 3}\frac{x^2+\alpha x +\beta}{x^2-8x+15}&=\lim_{x\to 3}\frac{(x-3)^2}{(x-3)(x-5)}\\[3pt] &=\lim_{x\to 3} \frac{x-3}{x-5}\\[3pt] &= 0 \end{align*}
• $f$ is continuous at $5$ iff $\lim_{x\to 5} f(x)=0$, iff $x^2+\alpha x+\beta =(x-5)^2$, cause in such a case \begin{align*} \lim_{x\to 5}\frac{x^2+\alpha x +\beta}{x^2-8x+15}&=\lim_{x\to 5}\frac{(x-5)^2}{(x-3)(x-5)}\\[3pt] &=\lim_{x\to 5} \frac{x-5}{x-3}\\[3pt] &= 0 \end{align*}

Therefore, we conclude the folowing: $f$ is discontinuous

• Only at $3$ if $\alpha = -10$ and $\beta = 25$

• Only at $5$ if $\alpha = -6$ and $\beta = 9$

• In both points $3$ and $5$ if $(-10,25)\neq (\alpha, \beta) \neq (-6, 9)$.