Refine to make continuos f(x) = { x+5 if x -2 I already know that it is a removable discontinuity, but I don't know how to go about redefining to make the function continuous.

Question

Refine to make continuos

f(x) = { x+5 if x<-2
             2 if x=-2
            1-x if x > -2

I already know that it is a removable discontinuity, but I don't know how to go about redefining to make the function continuous.

turingtest · Answer

the function is continuous at x=a iff$$\lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=f(a)$$

anonymous · Answer

if lim x->a f(x) exists but, lim x-> f(x) doesn't equal f(a) then we can redefine f

turingtest · Answer

right, so what is$$\lim_{x\to -2}f(x)$$?

anonymous · Answer

it's a removable discontinuity even though the function is continuos at x=-2

anonymous · Answer

=3

turingtest · Answer

well... then what should we define f(-2) to be?

anonymous · Answer

f(-2) = 2

turingtest · Answer

but we want$$\lim_{x\to-1}f(x)=f(-2)$$which it would be if f(-2)=?

turingtest · Answer

typo$$\lim_{x\to-2}f(x)=f(-2)$$

anonymous · Answer

-2?

turingtest · Answer

we want$$\lim_{x\to-2}f(x)=f(-2)$$what is$$\lim_{x\to-2}f(x)$$?

anonymous · Answer

=3

turingtest · Answer

right, so we want to make f(-2) have that value as well so that$$\lim_{x\to-2}f(x)=f(-2)$$savvy?
so we redefine$$f(-2)=?$$

anonymous · Answer

so since f(-2)=2, we want it to equal 3 so it will be continuous everywhere?

turingtest · Answer

yep :)

anonymous · Answer

now how do i go about doing that? :P

anonymous · Answer

this is the part that confuses me.

turingtest · Answer

simply by stating so

turingtest · Answer

redefine f(-2)=3

turingtest · Answer

that's all

anonymous · Answer

so i state that it is a removable discontinuity that is redefined f(-2)=3

i don't need to do anymore calculations?

turingtest · Answer

nope

anonymous · Answer

there is no way to actually redefine the function to make it =3?

turingtest · Answer

it is a step function
your job is to redefine it such that it is continuous at x=-2
by redefining f(-2)=3 you have solved the problem
there is no sorcery to be done here, just redefining the function and your done :)

anonymous · Answer

okay, thank-you.

if i have f(x)=x-2/|x-1| 
how do i determine the points that it is discontinuous at?

turingtest · Answer

like most rational functions, you want to look out for when the denominator is zero

anonymous · Answer

i know how to state the absolute value in terms of negative and positive.

anonymous · Answer

the denominator is zero when x=1

turingtest · Answer

right

turingtest · Answer

let me see if there are any more...

turingtest · Answer

nah that's all

anonymous · Answer

when i preform the limit, i get an undefined answer

turingtest · Answer

yes, the limit as x->1 is undefined

anonymous · Answer

i don't know how i would determine the points where it is discontinuous

turingtest · Answer

you already did, x=1
that's all
I think you are over-thinking these problems :P

turingtest · Answer

points where the function is undefined or not equal to the limit are discontinuous

anonymous · Answer

would that be a jump discontinuity? and i just don't understand calculus haha.

turingtest · Answer

no, a jump discontinuity is like|dw:1349292853855:dw|notice the function is defined everywhere here, but the limit from the right at the jump point is different from the limit from the left
that is a jump discontinuity

turingtest · Answer

in you equation you have the function completely undefined at x=1, so that is an "essential" discontinuity

turingtest · Answer

your*

anonymous · Answer

is that the same as an infinite discontinuity?

turingtest · Answer

yes, my terminology may be a bit weak, sorry

anonymous · Answer

but f(x)=x-2/|x-2| is a jump disconuity correct?

turingtest · Answer

yes, nice job

anonymous · Answer

is there anyway to redifine an inaqfinIT

anonymous · Answer

an infinity limit *
sorry im babysitting haha, she wants to play with the keyboard.

anonymous · Answer

infinite discontinuity **

turingtest · Answer

haha, no that is why I used the term "essential" discontinuity

turingtest · Answer

because it is essential to the function, and no value that you could give f(1) would make (x-2)/|x-1| continuous at x=1

anonymous · Answer

ah i see.

turingtest · Answer

the function approaches -infinity as x->1 so any finite value would not solve the problem

anonymous · Answer

for f(x)=x-2/|x-2|
do i preform the limit as x->2 or as x->1?

turingtest · Answer

you want to check the limits at the undefined points
is this thing undefined at x=1 ?

anonymous · Answer

no it is undefined at 2

turingtest · Answer

right, so that is all you need to check

anonymous · Answer

i end up with undefined numbers the same as when i did it for x-1/|x-1| but i know i am supposed to end up with a jump discontunity

turingtest · Answer

wait the earlier problem was (x-2)/|x-1| right?

anonymous · Answer

yes thats what i meant

turingtest · Answer

okay, so for (x-2)/|x-2| think about the possible values of the function
if x<2 what is the value of the function?

anonymous · Answer

-1

turingtest · Answer

right, and since it is the same value everywhere up until x=2 that means that$$\lim_{x\to2^-}f(x)=-1$$what about for x>2?

anonymous · Answer

1

turingtest · Answer

and so the limit from the right is...?

anonymous · Answer

1

turingtest · Answer

right

turingtest · Answer

so$$\lim_{x\to2^-}f(x)\neq\lim_{x\to2^+}f(x)$$though both limits do exist (contrast this with the infinite discontinuity in which the limits don't exist at all)
because the left and right limits both exist, but are not the same, this is a jump discontinuity

anonymous · Answer

and by stating that i have redefined it?

turingtest · Answer

if the left and right limits are not equal, then the limit does not exist and the discontinuity is essential
it may be a jump discontinuity (if the left and right limits exist but are different)
or it may be an infinite discontinuity (if either the left or right hand limit are undefined)