Question

what are the removable discontinuities of the following function? f(x) = (x^2 - 36)/(x^3 - 36x) o x = -6, x = 0, and x = 6 o x = -6 and x = 6 o x = 0 and x = -6 o x = 0 and x = 6

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-36=(x + 6)(x - 6)$ using 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 $x
eq - 6,6$). The simplified function is $f(x)=\frac{1}{x}$ for $x
eq - 6,6,0$. Removable discontinuities occur where the original function is undefined but the limit exists. The original function $f(x)=\frac{x^{2}-36}{x^{3}-36x}$ is undefined at $x=-6,0,6$. After simplification, the factors $(x + 6)$ and $(x - 6)$ that were causing the "hole" - like discontinuities can be removed.

Answer:

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