continuity of a function on intervals

**roguetrader** · January 31st 2010, 12:17 PM

hi guys
2nd year uni student here...havent taken math 30 pure for 3 years so i am regretting theh calculus i am doing, even though it may seem easy to many of you:
2x + x^2/3 on [-1,1]

...can someone walk me through a question like this and explain it in simpleton terms thanks

**VonNemo19** · January 31st 2010, 12:31 PM

Originally Posted by roguetrader

hi guys
2nd year uni student here...havent taken math 30 pure for 3 years so i am regretting theh calculus i am doing, even though it may seem easy to many of you:
2x + x^2/3 on [-1,1]

...can someone walk me through a question like this and explain it in simpleton terms thanks

What?

**roguetrader** · January 31st 2010, 02:50 PM

Originally Posted by VonNemo19

What?

is the function continuous on the given interval
2x + x^(2/3) on [-1,1]

**Plato** · January 31st 2010, 03:09 PM

Originally Posted by roguetrader

is the function continuous on the given interval
2x + x^(2/3) on [-1,1]

The function $2x+\sqrt[3]{{x^2 }}$ is continuous on every interval.

**roguetrader** · January 31st 2010, 03:42 PM

Originally Posted by Plato

The function $2x+\sqrt[3]{{x^2 }}$ is continuous on every interval.

why? can someone run me through how to get that solution?

**Krizalid** · January 31st 2010, 03:45 PM

A well defined function on its domain is continuous there.

**Plato** · January 31st 2010, 03:51 PM

Originally Posted by roguetrader

why? can someone run me through how to get that solution?

The answer to that question is: probably not.
Unless you can tell us what makes a function discontinuous.
Can you tell us that? If not I doubt you will follow any explication.

**VonNemo19** · January 31st 2010, 03:51 PM

Originally Posted by roguetrader

why? can someone run me through how to get that solution?

Check for differentiability.

Also, have a look at the definition of continuity.

We say A function $f(x)$ is continuos on an interval $[a,b]$ if

$f(x)$ is defined on $[a,b]$

$\lim_{x\to{c}}f(x)=f(c)$ for all $c\in(a,b)$

and

$\lim_{x\to{c}}f(x)$ Exists for all $c\in(a,b)$

**Plato** · January 31st 2010, 03:59 PM

Originally Posted by Krizalid

A well defined function on its domain is continuous there.

Surely that depends upon what one means by ‘well defined’.
I certainly consider $f(x) = \left\{ {\begin{array}{rl}<br /> {x - 1,} & {x < 1} \\<br /> {x^2 ,} & {x \geqslant 1} \\ \end{array} } \right.$ well defined on $[-1,1]$ .

**HallsofIvy** · February 1st 2010, 03:38 AM

Originally Posted by Krizalid

A well defined function on its domain is continuous there.

Ouch! I would say that "f(x)= 1 if $x\le 0$ , -1 if x> 0" is "well defined" but certainly not continuous on its domain!

It is true that our ways of writing functions have developed so that any function we are able to write as a single formula is continuous wherever it is defined but that an artifact of the way we write formulas and has nothing to do with being "well defined".

**Krizalid** · February 1st 2010, 06:13 AM

ahhh, i think that's not what i meant to say, i mean if $\text{Dom}f=\mathbb R$ then $f$ is continuous on $\mathbb R.$

that's what i said about "well defined."

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