Can somebody help me with continuity? :(

sure name a problem. one definition is \[f\] is continuous at a point a if \[\lim_{x\rightarrow a}f(x)=f(a)\]

intuitively it means f is continuous on an interval if when you graph the function over that interval you don't have to lift your pencil

Find a value for a so that the function f(x)={4-x^2, x<-1 and ax^2-1, x>=-1 is continuous.

ok got it

Lol yes, I actually have specific problems that I don't get. I understand the gist of continuity..

let me show off by writing it in latex, but the simple idea is plug in -1 into both expressions and set them equal

What's Latex? Lol

there how it that?

It is ax^2 if x>= -1

\[f(x) = \left\{\begin{array}{rcc} 4-x^2 & \text{if} & x <-1 \\ ax^2& \text{if} & > -1 \end{array} \right.\]

ok fine. this problem actually has very little to do with continuity. the idea is that you want to function value on the left to equal the value on the right

so although the idea my be confusing, the solution is easy. replace x by -1 on both sides

so it will 3 = a-1? a=4?

i .e. put \[4-(-1)^2=a(-1)^2\] \[4-1=a\] \[a=3\] tada

lets be careful because i see what you wrote above and it is not exactly what we have

Lol ax^2 - 1...

is it \[ax^2\] if x> -1 or is it \[ax^2-1\]?

ok well you get the idea.

\[ax ^{2} - 1\]

plug in -1, solve for a

Ok, thanks so much :)

yw

1 more problem? ><

Give a formula for the extended function that is continuous at the indicated point. f(x)=(x^3-4x^2-11x+30)/(x^2-4); x=2

post or just ask

got it

I posted, but no one answered Lol

easy enough again. factor and cancel

How do I factor the numerator? Lol I forgot..

you get \[\frac{x^3-4x^2-11x+30}{x^2-4}=\frac{x^3-4x^2-11x+30}{(x-2)(x+2)}\]

let me give you a huge huge hint, because this will come up frequently. you are asked to make it continuous at \[x=2\] right?

Yes

and the reason this is not continuous at 2 is because of the factor of \[x-2\] in the denominator right?

so if you have any hope of making it continuous at x = 2 then the numerator MUST factor as \[(x-2)\times \text{something}\]

Yes, but won't be it not continuous at x=-2 as well? But right now, it will be an infinite discontinuity right? How can I fix that...

you cannot fix an infinite discontinuity, only a removable one

that is why it is called "removable" because you can remove it.

but i hope my suggestion was clear, it MUST factor as \[(x-3)\times \text{stuff}\]otherwise you cannot cancel the factors of x - 2 from top and bottom

so your real job is just to factor the numerator, cancel the x - 2 and write down your answer. you get \[\frac{(x-2)(x^2-2x-15)}{(x+2)(x-2)}\]

for a final answer of \[\frac{x^2-2x-15}{x+2}\]

So I have to break x^3-4x^2 - 11x + 30 so that I can break it down to (x+2) X something?

exactly!

if it does not factor that way, you cannot do the problem

you do it by division or by cheating. your choice

Oh I see, so now this function can be continuous at x=2.. Ohh :O

Cheating? How do you do it by cheating?

RIGHT! and if you could not cancel ( i know i am repeating myself) you can't do it

cheat like this http://www.wolframalpha.com/input/?i=%28x^3-4x^2-11x%2B30%29%2F%28x^2-4%29

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