Question

let $f(x)$ be defined as follows.
$f(x)=\frac{x^{2}-1}{x^{2}-6x + 5}$
give the interval(s) on which the function is continuous. use interval notation, or enter dne if no such intervals exist. include the union symbol when entering multiple intervals.
interval(s) of continuity:

Explanation:

Step1: Factor the numerator and denominator

$x^{2}-1=(x + 1)(x - 1)$ and $x^{2}-6x + 5=(x - 1)(x - 5)$
So, $f(x)=\frac{(x + 1)(x - 1)}{(x - 1)(x - 5)}$, $x
eq1,5$. After canceling out $(x - 1)$ (for $x
eq1$), $f(x)=\frac{x + 1}{x - 5}$, $x
eq1,5$.

Step2: Find the points of discontinuity

The function is discontinuous where the denominator is zero. For $y=\frac{x + 1}{x - 5}$, the denominator $x-5 = 0$ when $x = 5$, and we also had a removable - discontinuity at $x = 1$ from the original form.

Step3: Determine the intervals of continuity

The function is continuous on the intervals $(-\infty,1)\cup(1,5)\cup(5,\infty)$.

Answer:

$(-\infty,1)\cup(1,5)\cup(5,\infty)$