Question

[1 , infinity) except 1 How would you represent this new domain into simpler interval form?

Answer 1

\[1\le x <\infty \]

Answer 2

but interval notation is what u have already written

Answer 3

Except" 1 "

Answer 4

F(x) = 2 + (sqrt x -1) G(x) = sqrt (x -1)

Answer 5

Find the domain of f/g

Answer 6

ok .........then put an open interval at one
\[(1,\infty)\]

Answer 7

But I kind get confused because of that denominator which theorem will I follow ?

Answer 8

n/ Sqrt (x-1) > Which theorem will I follow is it the not equal to zero or greater than zero

Answer 9

he domain of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or value for each member of the domain.[1] The set of values the function may take is termed the range of the function.

Answer 10

for \[\sqrt{x-1}\] the domain is \[x-1\ge 0\] but since in this case sqrt(x-1) is in denominator,so it cannot be equal to zero.ie.domain is x-1>0
x>1
\[(1,\infty)\]
@Reyjuanx10

Answer 11

So if it is in the denominator automaticilly it must be not equal to zero even if it has square root?

Answer 12

Can I apply rationalizing ?

Answer 13

ya .........i ll give an example::
say the function is 1/(x-2)
now domain is the possible set of values x can have.........
now x cannot be equal to 2 as if x=2 the value of function will be infinite which is not possible....
so domain in this case will be all real numbers except 2
...
square implies that the nos.....have to be positive....right

Answer 14

this was the reply of ur denominator question

Answer 15

Got it . Thanks !

Answer 16

ur welcome.....