Question

This is a Calculus question:
I need help understanding algebraic functions.
Function given below.
Things in need to find: domain; points of discontinuity and type. Please Help! Test tomorrow!!!

Answer 1

(Square root) of -/x^2 -9
Divided by. X+3

Answer 2

i am going to guess it is
\[f(x)=\frac{\sqrt{x^2-9}}{x+3}\] but i could be wrong

Answer 3

Correct thank you...that thing wouldn't come up for me :)

Answer 4

@satellite73

Answer 5

@Hero

Answer 6

ok
domain first

Answer 7

it should be pretty clear right away before we worry about the numerator that \(x\neq -3\) right?

Answer 8

Alright

Answer 9

and since you cannot take the square root of a negative number, we also have to solve
\[x^2-9\geq 0\] as well
do you know how to solve that for \(x\) ?

Answer 10

(X-3)(x+3)>(or equal to) 0

Answer 11

xactly
you know the zeros are at -3 and 3, and since it is a parabola that opens up
(i mean \(y=x^2-9\) is the parabola) then it must be positive outsides the zeros and negative between them
you want 'positive" so it will be \[(-\infty,-3]\cup [3,\infty)\]but now we have to be careful

Answer 12

since we have excluded \(-3\) from the domain, the real answer is
\[(-\infty, -3)\cup [3,\infty)\]

Answer 13

Could you write that answer in another way ?

Answer 14

sure
\(x<-3\) or \(x\geq 3\)

Answer 15

Ok :)

Answer 16

So that is the domain, so how do you find the points of discontinuity?

Answer 17

it is discontinuous at the end points of the domain

Answer 18

at 3 it is discontinuous because the function does not exist to the left of 3

Answer 19

at -3 it is discontinuous for the same reason, but also because as x goes to -3 from the left, the whole thing goes to \(-\infty\)

Answer 20

3 would be a nonremovable point of discon. And -3 would be a removable pt. of discon.?

Answer 21

@satellite73

Answer 22

neither are removable

Answer 23

removable basically means you have a hole in the graph
you can mostly remove it by factoring and cancelling to get an equivalent expression
that is not true in this case

Answer 24

So the lim f(x)= -infinity
X->3-
Lim f(x)= +infinity
X->3+

Answer 25

first one is true second one is not

Answer 26

\[\lim_{x\to 3^+}f(x)=0\]

Answer 27

then numerator goes to zero, the denominator goes to 6, and \(\frac{0}{6}=0\)

Answer 28

Oh ok! Thank you!,

Answer 29

yw