continuity functions

You can use derivates of the original function to find the critical numbers of the function, which will help you graph the equation. You can find the concavity and whether it is increasing or decreasing, which will help you graph.

I do not have my graphing calculator with me, but if you do, you could just plug the equation into it to find whatever information you need.

but I'm.supposed to use the definitions aren't i

Posted from TSR Mobile

I am not sure what you mean by definitions (I am American by the way, that may why I don't get it)

having trouble with all of these but I had a go at all of them. I doubt they're correct . please help.
I get the feeling that I'm meant to use the epsilon delta stuff on the first one. i really don't think its continuous since there's always gonna be jumps.

Posted from TSR Mobile

as the limit point in the last question is finite, and the actual limit is 0/0 - then why not (if you are allowed) use L`Hopital`s rule?

(it`s only going to be a one-sided limit, since x> -1)

the 1`s disappear, and you have a number times a numerator and denominator in different powers of (x+1) you can then use the rule of indices on to simplify. Take the limit as x approaches 0 and find your point.

Then you can define the continuous extension on f(x) as a piecewise (continuous) function of being: (what you get) at x=0, the f(x) you`re given for all x>-1)

(the 1st function IS piecewise continuous, because the value of the function is defined at x=-1)

say, alpha =3, we have (see graph)

(that is the continuous extension of 1/(x+1)^2 on R)

Sincere apologies cooldudeman! ignore my post as far as Q1) goes - it`s utter C**p! (I will edit the c**p out of it, though!)

Again - apologies. (the 1st function is discontinuous at x=-1 because although from left and right it approaches the same limit - it`s not finite - and it is as you say undefined at x= -1)

Sorry - for now - maybe someone can come up with a better solution - L`Hopital is the only thing i`ve tried that`s worked (but, then again, i`m Dog-tired!)

the key point about L`Hopital is that if you encounter, in rational functions, infinite or undefined limits of , for example, the form $\frac{0}{0}$, you can use it.

to use it:

1) differentiate the numerator and denominator just as they are, so you`d get:

$\displaystyle \frac{\frac{1}{2}(x+1)^{-1/2}}{\frac{1}{3}(x+1)^{-2/3}}$

re-arrange this into a single power of (x+1) times a number, and take the limit as x->0

(i`m off to bed right now - cream crackered!)

Another way, kind of complicated, is to rationalise it with a little trick.

to begin with, multiply the given rational function, top and bottom, by something which will turn the denominator into a difference of 2 cubes (this will simplify the denominator unbelievably!

namely, multiply top and bottom by

$(x+1)^{2/3}+1^{2}+(x+1)^{1/3}$

(because, believe it or not, this is the "rest of" the factorisation of just $x$ for a difference of 2 cubes!

you should then get:

$\displaystyle (\frac{\sqrt{x+1}-1}{x})(1+(x+1)^{1/3}+(x+1)^{2/3})$

you can then just rationalise the first bit in brackets, by multiplying top and bottom by:

$\sqrt{x+1}+1$ - take the limit, and Bob`s your 3/2`s!!

(EDITED)

(it`s from the general formula for $a^{n}-1=(a-1)(a^{n-1}+a^{n-2}+...+a^{n-(n-1)}+1)$ where here, $a=(x+1)^{1/3}$ )

so, you`d define, in the function, $f(0)=\frac{3}{2}$

All of the above answers assume results that you have not yet proven, and hence are useless to you.


1) Graph sketching should be easy at this point - just mark on the limits (and direction of approach), asymptotes (and behaviour at them), critical values (if any), axis intercepts (if any) and anything else important. You can do this purely by inspection. For the continuity part, just look at the function and you should be able to tell which way you're going to want to prove (if not, look at your sketch and it should be immediately obvious). Once you know which way you're going, just use the definition (or its negation, or something that you've proven equivalent to one of the above).
2) By inspection, it should be clear where the discontinuities are (also, that's a dreadful question, as it doesn't state the domain of the function, but never mind). Just prove that each is a discontinuity (by the negation of the definition or something equivalent to it), and prove that it is continuous everywhere else (by the definition, sum/product/quotient/etc. rules, etc.). For the types part, you'll have to tell us exactly what "types" you are using.
3) Use the limit definition of continuity, if you have it. If not, just take the limit near zero to guess your $f(0)$, and use a definition that you do have to prove that $f$ is continuous at zero with that $f(0)$, and no other $f(0)$.

this is what I got now.
still not sure about any of them.
for 2) I have no idea how you prove them because it seems obvious what I wrote about the LH and RH limits being the same. I was just using the definitions of the types of discontinuity

for 3) really sorry but I have no idea what you meant. and I don't think I have this limit definition of continuity as I went through the lecture notes again.

Posted from TSR Mobile

I can think of at least four mutually exclusive ways of defining "types" of discontinuity, just off the top of my head. I can't help you until you tell me which one you're using.



The limit definition of continuity is: $f:[a,b]\to\mathbb{R}$ (or on whatever other sensible domain you feel like using) is continuous at $x_0 \in [a,b]$ iff $\lim_{x\to x_0} f(x) = f(x_0)$. If you don't have it, you can prove it excessively easily.