Having trouble with Continuity crap.

[25thhour]

This is the first problem

f(x) { x + c , if x < 2
{ cx^2 + 1 , if x >= 2

Find the values of the constant C so that f(x) is continous at x = 2

So what I know is that for a function to be continous you need 3 things.

f(x) must be defined, the limit must exist, and f(x) must equal the limit.

First step I did was make both equations equal

x + c = cx^2 + 1

Next, I subbed x = 2 into the equation

2 + c = 4c + 1

Then I found c

1 = 3c
c = 1/3

Now I subbed C back in to the equations to get

x + 1/3
and
(1/3)x^2 + 1

I then attempted to check points greater than or equal to two into the 2nd function (2) and points less than 2 (0) into the first function. (This is where I don't know what I am doing)

I get 7/3 for the 2nd one
and 1/3 for the first.

They dont equal so this can't work... I am confused.

[25thhour]

Re: Having trouble with Continuity crap.

Oh man, I just got it... Fuckin durp...

Okay, Lol.

Instead of testing points. I made sure the function was defined at 2 (it was 7/3)
Then I took the limits at 2 from the negative and the positive, they both equaled 7/3.

yay it worked


C = 1/3

EDIT:

I do have one more problem I will post in a bit

[iCeCuBEz v2]

Re: Having trouble with Continuity crap.

gj

[25thhour]

Re: Having trouble with Continuity crap.

I do have one more problem that is just limits at infinity:

lim x-> -inf [sqrt(x^2 + 4) - cubedroot(8x^3 + x - 15)] / (4x - 5)

so what I got was -1/4 because since when you simplify the highest power is x^1 you take the coefficients 1 - 2 = -1 and put it over the bottom 4.

I would like someone to check this please as wolfram gets a wonky ass answer.

[Zapmeister]

Re: Having trouble with Continuity crap.

copying and pasting your thingy into wolfram alpha gives -3/4, and if you tell it to substitute x= -1000000 or something it gives a close enough answer.

the discrepancy you're having is to do with the signs in the term with the sqrt in it. that's because sqrt(x^2) isn't x, it's |x|. so when you "take the highest power" you get sqrt(x^2 + 4) / (4x-5) --> |x| / (4x) --> -1/4 as x--> -inf.

alternatively if you start with sqrt(x^2 + 4) / (4x-5) and divide top and bottom by |x| you get sqrt(1 + 4x^-2) / (4 sgn[x] -5/|x|) which goes to -1/4 as x --> -inf.

you don't need to worry about anything like this for the cubed root term because the cube root function is one-to-one and cuberoot(x^3) = x for all x, so that bit is fine.

[benguino]

Re: Having trouble with Continuity crap.

@25th: I sent you an FFR PM with some help for that second problem (sometimes I'm self-conscious about posting math in the ffr forums and I decide I don't want to)

[25thhour]

Re: Having trouble with Continuity crap.

Ben I never received the PM.

Im confused about the absolute value? Why does it change to -3/4? Am I just accidently switching signs? Should it actually be -1? because it is the absolute value of x x being an absolute value?

[Zapmeister]

Re: Having trouble with Continuity crap.

ok, let's put it this way.

lim x-> -inf [sqrt(x^2 + 4) - cubedroot(8x^3 + x - 15)] / (4x - 5)

let's split it up into two bits.

lim x-> -inf [sqrt(x^2 + 4)] / (4x - 5) - lim x-> -inf [cubedroot(8x^3 + x - 15)] / (4x - 5)

look at the first bit, lim x-> -inf [sqrt(x^2 + 4)] / (4x - 5)

what happens if x becomes *really* negative? try x = -million. then sqrt(x^2+4) = sqrt (trillion and 4) = like a million, and (4x - 5) is like -4 million, so you suspect the limit of that bit should be -1/4.

what i think you're trying to write is [sqrt(x^2 + 4)] --> x, and (4x - 5) --> 4x, and you get x / 4x = 1/4. what happened to the minus sign?

the problem here is that sqrt(x^2) = |x|, not x. if you square a big negative number and square root it, you get the positive of what you started with, since square root is defined to always be positive. for instance, if you had x=3, sqrt(x^2)=3. but if you had x=-3, sqrt(x^2)=3 also.

so when you "take the highest power" in the numerator, you should do [sqrt(x^2 + 4)] --> |x| instead. then, the denominator goes to 4x, and what you have is |x| / 4x. when you have x negative, |x|/x is -1, so this is -1/4.

lim x-> -inf [sqrt(x^2 + 4)] / (4x - 5) = -1/4

for the second term, cubedroot(8x^3 + x - 15) is close to cubedroot(8x^3), or 2 cuberoot(x^3) for large x. here you don't have to do that again, because cuberoot(x^3) = x for all x. this is because each real number always has exactly one cube root.

let me know if you're still confused

[25thhour]

Re: Having trouble with Continuity crap.

ok, let's put it this way.

lim x-> -inf [sqrt(x^2 + 4) - cubedroot(8x^3 + x - 15)] / (4x - 5)

let's split it up into two bits.

lim x-> -inf [sqrt(x^2 + 4)] / (4x - 5) - lim x-> -inf [cubedroot(8x^3 + x - 15)] / (4x - 5)

look at the first bit, lim x-> -inf [sqrt(x^2 + 4)] / (4x - 5)

what happens if x becomes *really* negative? try x = -million. then sqrt(x^2+4) = sqrt (trillion and 4) = like a million, and (4x - 5) is like -4 million, so you suspect the limit of that bit should be -1/4.

what i think you're trying to write is [sqrt(x^2 + 4)] --> x, and (4x - 5) --> 4x, and you get x / 4x = 1/4. what happened to the minus sign?

the problem here is that sqrt(x^2) = |x|, not x. if you square a big negative number and square root it, you get the positive of what you started with, since square root is defined to always be positive. for instance, if you had x=3, sqrt(x^2)=3. but if you had x=-3, sqrt(x^2)=3 also.

so when you "take the highest power" in the numerator, you should do [sqrt(x^2 + 4)] --> |x| instead. then, the denominator goes to 4x, and what you have is |x| / 4x. when you have x negative, |x|/x is -1, so this is -1/4.

lim x-> -inf [sqrt(x^2 + 4)] / (4x - 5) = -1/4

for the second term, cubedroot(8x^3 + x - 15) is close to cubedroot(8x^3), or 2 cuberoot(x^3) for large x. here you don't have to do that again, because cuberoot(x^3) = x for all x. this is because each real number always has exactly one cube root.

let me know if you're still confused

I think I remember covering it in class now, something about when you squareroot x^2 you get like -|x| when what you were supposed to substitute in was negative.

[Arkuski]

Re: Having trouble with Continuity crap.

Now prove continuity using epsilon-delta notation