Question

Prove √ ̅(x+3) (square root of the quantity x + 3) is continuous at all points in its domain.

Answer 1

is this for a calculus class?

Answer 2

yeah. :)

Answer 3

i think has something to do with showing that for all a>= -3
\[\lim_{x \rightarrow a}\sqrt{x+3} = \sqrt{a+3}\]

Answer 4

yeah, i have some idea of how to do it. But I hope someone can show me clearly.

Answer 5

i forget how to to prove it for every point,
you can pick certain points and show its continuous at that point

Answer 6

ahh you can also differentiate
if a function is differentiable it must be continuous

Answer 7

\[\frac{d}{dx}\sqrt{x+3} = \frac{1}{2\sqrt{x+3}}\]

Answer 8

f'(x) is defined for all points x>-3
thus f(x) is continuous at all points x>-3

Answer 9

\[\lim_{x \rightarrow x _{0}} f(x) = \lim_{x \rightarrow x _{0}} \sqrt{x+3}=\sqrt{x _{0} + 3}\]
with all \[x _{0}\] \[\in\] [-3,\[\infty\]]

That's what the proof says, but I have no idea how it went from \[\lim_{x \rightarrow x _{0}} \sqrt{x+3}\] to \[\sqrt{x _{0} + 3}\]

Answer 10

when you evaluate a limit, you replace the variable with whatever it approaches
thats how it became \[\sqrt{x _{0} +3}\]

Answer 11

well, in order for you to do that, you have to be able to prove that the function is continuous at that point first, which we haven't done yet.

Answer 12

You can't differentiate to prove continuity, in order to differentiate you must first know that the function is differentiable, which means you must know that the function is continuous, among other things.
Limits is a way to prove this, if you want a really strict proof you should use an epsilon-delta argument.

You could also use that if f(x) and g(x) are continuous, then so is f(g(x)).

I hope that all made sense.

Answer 13

yep thats the way to go...epsilon-delta limit proof
sorry for the bad advice...proofs are not my speciality