what are the removable discontinuities of the following function?\nf(x)=\frac{x^{2}-36}{x^{3}-36x}\n(x =…

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-36=(x + 6)(x - 6)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. The denominator $x^{3}-36x=x(x^{2}-36)=x(x + 6)(x - 6)$. So, $f(x)=\frac{(x + 6)(x - 6)}{x(x + 6)(x - 6)}$.

Step2: Simplify the function

Cancel out the common factors $(x + 6)$ and $(x - 6)$ (for $x\neq\pm6$ and $x\neq0$). The simplified function is $f(x)=\frac{1}{x}$ for $x\neq\pm6,0$.

Step3: Identify removable discontinuities

Removable discontinuities occur where the original function is undefined but the simplified function is defined. The original function $f(x)=\frac{x^{2}-36}{x^{3}-36x}$ is undefined at $x=-6,x = 0,x = 6$. But since we can cancel out the factors $(x + 6)$ and $(x - 6)$, the discontinuities at $x=-6$ and $x = 6$ are removable.

Answer:

B. $x=-6$ and $x = 6$