Question

determine where the following function is continuous....

Answer 1

\[y=\sqrt{2x+3}\]

Answer 2

it's continuous on (-3/2, infinity)

Answer 3

thanks for the answer??? but how did you get that?

Answer 4

\(\large y=\sqrt x \) is continuous for x > 0, so all you need to do is solve 2x+3 > 0....

Answer 5

@ByteMe Sure, but isn't it \(x\geq 0\). There is nothing wrong with \(\sqrt{0}\).

Answer 6

Or maybe it's that it isn't continuous?

Answer 7

hmmm... i think ur you're right...

Answer 8

But \[\Large \lim_{x\rightarrow \frac{-3}{2}} \sqrt{2x+3} = 0 \]And\[f\left(\frac{-3}{2}\right)=0\]So it is continuous.

Answer 9

well, the definition of continuity is defined as LEFT and RIGHT limits have to be the same.... but in this case, the left limit does not exist because the domain restriction.....

Answer 10

Yeah, that's the part that is tricky is that the limit should be defined on both sides...

Answer 11

lol... that's why i try to avoid these questions.... :)

Answer 12

But if they asked for an epsilon... couldn't we give them a delta?

Answer 13

yes.... give any delta and you should be able to find the epsilon that coincides..

Answer 14

Yeah, my limit up there probably isn't valid because it assumes it is continuous at 0.

Answer 15

so either the answer is:
[-3/2, infinity) or (-3/2, infinity)

take your pick asker... sorry i'm not much help....

Answer 16

@Hero @hartnn @mathslover @ganeshie8 What do you think?!