Question

Classify the discontinuities (if any) for the given function:f(x)={(x+4)/(x^2-3x-28) x not equal 4, x not equal 7
{-1/11 otherwise

Answer 1

same as
\[\frac{1}{x-7}\]

Answer 2

the discontinuity at 7 is infinite, i.e.it has a vertical asymptote there

Answer 3

why not removable at x=-4?

Answer 4

the discontinuity at-4 is evidently removable, because i just removed it.

Answer 5

haha, let me try again :-)

Answer 6

it is removable at -4, but not at 7

Answer 7

can you pls tell me how it is removable at -4?

Answer 8

we removed it when replacing
\[\frac{(x+4)}{(x^2-3x-28)}\]
by
\[\frac{1}{x-7}\]

Answer 9

I got (x+4)/((x-7)(x+4))

Answer 10

doesn't that imply a hole at x=-4?

Answer 11

more precisely, your function was
\[\frac{x+4}{x^2-3x-28}=\frac{1}{x-7}\]

Answer 12

yes it has a hole at -4

Answer 13

but if you define it to be
\[-\frac{1}{11}\] the hole is filled up. that is why it is called "removable" because you have just removed it

Answer 14

oh!!! I never knew that! thank you so much!

Answer 15

it doesn;t matter that it says 'otherwise' by -1/11?

Answer 16

yw. math often names thinks what they are. not always of course. but in this case "removable' means you can remove it, "jump" means there is a jump, and "infinite" means it goes to infinity

Answer 17

ahh "otherwise"

Answer 18

interesting

Answer 19

otherwise = non-integers (fractions, decimals, irrational numbers,...)

Answer 20

that means you have defined
\[f(x)=\frac{x+4}{x^2-3x-28}\] unless
\[x=-4,7\]
and
\[f(-4)=-\frac{1}{11}\] also
\[f(7)=-\frac{1}{11}\]

Answer 21

now
\[f(-4)=-\frac{1}{11}\] fills the whole so removes the discontinuity there

Answer 22

however
\[f(7)=-\frac{1}{11}\] doesn't fill a hole because there is no hole there. function goes to infinity. i just defines the function there

Answer 23

how is f(7) = -1/11? when I plug in I do not get that?

Answer 24

wait

Answer 25

you can define a function any way you like

Answer 26

understand the filling of a hole now!

Answer 27

i can say
\[f(x)=x^2\] unless x = 2, and
\[f(2)=50\]

Answer 28

good!

Answer 29

filling in the hole means just that. plugging it up

Answer 30

so if the limit of my function is
\[-\frac{1}{11}\] at x = -4 i can define it to be
\[-\frac{1}{11}\] at -4 and that fills the hole

Answer 31

but don't forget i can define a function any way i like. i can say
\[f(x)=\frac{1}{x}\] and
\[f(0)=5\]

Answer 32

the fact that the formula
\[\frac{1}{x}\] has no meaning at x = 0 doesn't stop me from defining my function to be whatever i like at x = 0

Answer 33

so for f(7) - -1/11 you are not plugging in but rather defining

Answer 34

exactly!

Answer 35

*so for f(7) =-1/11 you are not plugging in but rather defining

Answer 36

right. just saying it.

Answer 37

that is rather awesome!

Answer 38

thanks!!!

Answer 39

that is one thing great about math. it is my function, i can define it as i choose

Answer 40

yw