Question

Can somebody help me with continuity? :(

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

ok got it

Answer 5

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

Answer 6

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

Answer 7

What's Latex? Lol

Answer 8

there how it that?

Answer 9

It is ax^2 if x>= -1

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Lol ax^2 - 1...

Answer 17

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

Answer 18

ok well you get the idea.

Answer 19

\[ax ^{2} - 1\]

Answer 20

plug in -1, solve for a

Answer 21

Ok, thanks so much :)

Answer 22

yw

Answer 23

1 more problem? ><

Answer 24

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

Answer 25

post or just ask

Answer 26

got it

Answer 27

I posted, but no one answered Lol

Answer 28

easy enough again. factor and cancel

Answer 29

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

Answer 30

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

Answer 31

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?

Answer 32

Yes

Answer 33

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

Answer 34

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

Answer 35

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...

Answer 36

you cannot fix an infinite discontinuity, only a removable one

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

exactly!

Answer 43

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

Answer 44

you do it by division or by cheating. your choice

Answer 45

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

Answer 46

Cheating? How do you do it by cheating?

Answer 47

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

Answer 48

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