Question

f(x)=sqrt(x-1) is either continuous from the right or continuous from the left at x=1. Determine which is true of the function and explain why one is true and the other false.

Answer 1

when is \(\sqrt x\) undefined?

Answer 2

(or at least not real)

Answer 3

at 0?

Answer 4

\(\sqrt0=0\) so no
when is \(\sqrt x\) not real, i.e. imaginary

Answer 5

?

Answer 6

I'm not sure. I have alot of trouble with this section

Answer 7

what is \(\sqrt{-1}\) ?

Answer 8

i

Answer 9

imaginary, right?
as in, not a real number
whenever anything under the radical sign is negative we get an imaginary answer, so the limit from the side on which the argument of the radical sign is negative does not exist, but the other does

Answer 10

the argument is negative when\[x-1<0\]so which limit does not exist?

Answer 11

the one at 0?

Answer 12

I don't think you understand the question
they are asking about the limit at x=1
they are asking whether the limit from the left or right does not exist, and why

Answer 13

so how can i solve this?

Answer 14

the limit at x=1 is 0

Answer 15

you cannot let the inside of a root be negative... that's why u say
"x-1>0"
so that the equation exists in real numbers... so in this case it will be
x>1 so "x must be greater than 1" and there u have your answer.... when numbers go up they are going to the right...

if the problem would have sayed :
f(x)=sqrt(1-x) then you would have

1-x>0 then
1>x so "x must be smaller than 1 att all timers

Answer 16

and actually what is inside the root can be 0 so at your excercise it should be
x-1>= 0