Question

Find the domain and range of the real function. f(x) = √x-1.

Answer 1

If the function is real, then you can't have an imaginary square root.
A square root is imaginary when the thing inside it is negative.

Answer 2

So, you only start with 0.
You can't have things such as\[\sqrt{-1}, \sqrt{-2}\cdots \]

Answer 3

So,\[ x - 1\ge0\]Solve that inequality. When you do, we'd sort the domain and range out too.

Answer 4

@ashna You there?

Answer 5

x^2 - 1 = 0

Answer 6

x - 1 > 0

Answer 7

What? Just solve the equality \(x - 1\ge 0\)

Answer 8

x > 1

Answer 9

inequality*

Answer 10

Yes.

Answer 11

what i did is wrong ?

Answer 12

So, \(x\) needs to be greater than or equal to 1. That's just natural numbers. Can be written as:\[\text{Domain:} x\in \mathbb{N} \]

Answer 13

No, it is correct.

Answer 14

okay

Answer 15

and range is also greater than or equal to zero right ?

Answer 16

Wait a second. Is the question like this?\[ f(x) = \sqrt{x} - 1\]

Answer 17

I was correct the first time. Never mind the second solution.

Answer 18

okay :)

Answer 19

Yes, @ashna
You are correct for the range too!

Answer 20

oh okay then .. tnx again :)

Answer 21

Can you please confirm if it is \(\sqrt{x}-1\) or \(\sqrt{x - 1}\)?

Answer 22

I have given the solution for the latter.

Answer 23

whole root x - 1