Question

What are the removable discontinuities of the following function?
f(x)=x^2-36/x^3-36x

Answer 1

Answer:

the removal discontinuity of the following function at x=-6 or x=6.

Step-by-step explanation:

$f(x)=\frac{x^2-36}{x^3-36x}$

$f(x)=\frac{(x-6)(x+6)}{x(x^2-36)}$

$f(x)=\frac{(x-6)(x+6)}{x(x-6)(x+6)}$

Since discontinuity are at x=0 ,6 or -6.

After cancelling, it leaves you with $f(x)=\frac{1}{x}$ for x≠0.

Therefore, x+6=0(or x=-6) , x-6=0(or x=6)  is a removal discontinuity of the following function.

Answer 2

Using vertical asymptotes, it is found that the function has removable discontinuities at $x = \pm 6$.

What are the vertical asymptotes of a function f(x)?

• The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.

• At these vertical asymptotes, the function is discontinuous.

• It a discontinuity can be factored, it is called removable.

In this problem, the function is given by:

$f(x) = \frac{x^2 - 36}{x^3 - 36x}$

Factoring it, we have:

$f(x) = \frac{x^2 - 36}{x(x^2 - 36)}$

Hence, the discontinuities are:

$x = 0$

$x^2 - 36 = 0$

Since the numerator also has a term $x^2 - 36 = 0$, this discontinuity can be factored and is removable, hence:

$x^2 - 36 = 0$

$x^2 = 36$

$x = \pm \sqrt{36}$

$x = \pm 6$

The function has removable discontinuities at $x = \pm 6$.

You can learn more about removable discontinuities at https://brainly.com/question/11598999