# Math Help - continuity of a function on intervals

1. ## 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

2. 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?

3. Originally Posted by VonNemo19
What?
is the function continuous on the given interval
2x + x^(2/3) on [-1,1]

4. 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.

5. 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?

6. A well defined function on its domain is continuous there.

7. 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.

8. 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 $c\in(a,b)$

9. 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 $[-1,1]$.

10. 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".

11. 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."