f(x) = 1/(5-(x^2+9)^1/2)((x^2+1)/((1-x)^4))^1/2

## How do I find all points where the following function is discontinuous?

# 2 Answers

A function is said to have a point of discontinuity at x = a or the graph of the function has a gap at x = a, if the original function is undefined for x = a, whereas the related rational expression of the function in simplest form is defined for x = a.

... it looks like the denominators are (5 - √(x^{2} + 9)) and √(1-x)^{4} then function is undefined if

**1.** 5 - √(x^{2} + 9) = 0 or **
2.** √(1-x)^{4} = 0

√(x^{2} + 9) = 5 x = 1

x^{2} + 9 = 5^{2}

x^{2} = 16

x = ± 4

The points of discontinuity are {- 4, 1, +4}

You want to consider where the denominator will equal zero.

There are two pieces to check.

(1-x) = 0 So, x = 1

5-sqrt(x^2+9)=0 Or, x= +-5

Graphing it can also help.

## Comments

Hi Savannah, can you, please, specify the denominator(s) of this function.

5-v(x^2+9) and v(1-x)^4 ... However, I'm a little confused with the second one because it is originally v((x^2+1)/(1-x)^4) ... So I don't know if that counts as a denominator by itself

Comment