# continuity of a function on intervals

• Jan 31st 2010, 12:17 PM
continuity of a function on intervals
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
• Jan 31st 2010, 12:31 PM
VonNemo19
Quote:

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?
• Jan 31st 2010, 02:50 PM
Quote:

Originally Posted by VonNemo19
What?

is the function continuous on the given interval
2x + x^(2/3) on [-1,1]
• Jan 31st 2010, 03:09 PM
Plato
Quote:

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.
• Jan 31st 2010, 03:42 PM
Quote:

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?
• Jan 31st 2010, 03:45 PM
Krizalid
A well defined function on its domain is continuous there.
• Jan 31st 2010, 03:51 PM
Plato
Quote:

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.
• Jan 31st 2010, 03:51 PM
VonNemo19
Quote:

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 $c\in(a,b)$
• Jan 31st 2010, 03:59 PM
Plato
Quote:

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 $f(x) = \left\{ {\begin{array}{rl}
{x - 1,} & {x < 1} \\
{x^2 ,} & {x \geqslant 1} \\ \end{array} } \right.$
well defined on $[-1,1]$.
• Feb 1st 2010, 03:38 AM
HallsofIvy
Quote:

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".
• Feb 1st 2010, 06:13 AM
Krizalid
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."