Question

Find the x-value of the removable discontinuity of the function. ​

Answer 1

Answer:

x=2

Step-by-step explanation:

$f(x)=\frac{x^2-4}{x^2-12x+20}$

Factor the numerator and denominator:

$f(x)=\frac{(x-2)(x+2)}{(x-2)(x-10)}$

We can remove (x-2) from the function:

$f(x)=\frac{x+2}{x-10}$

(x-2) is a removable discontinuity.

The x-value of x-2 is x=2.