let $f(x)$ be defined as follows.\n$f(x)=\frac{x^{2}-1}{x^{2}-6x + 5}$\ngive the interval(s) on which the…

let $f(x)$ be defined as follows.\n$f(x)=\frac{x^{2}-1}{x^{2}-6x + 5}$\ngive 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.\ninterval(s) of continuity:

let $f(x)$ be defined as follows.\n$f(x)=\frac{x^{2}-1}{x^{2}-6x + 5}$\ngive 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.\ninterval(s) of continuity:

Answer

Explanation:

Step1: Factor the numerator and denominator

$x^{2}-1=(x + 1)(x - 1)$ and $x^{2}-6x + 5=(x - 1)(x - 5)$ So, $f(x)=\frac{(x + 1)(x - 1)}{(x - 1)(x - 5)}$, $x\neq1,5$. After canceling out $(x - 1)$ (for $x\neq1$), $f(x)=\frac{x + 1}{x - 5}$, $x\neq1,5$.

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:

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