Question

f : R -> R : x|-> sqrt(x+1)

Q: which of the following is a well defined fucntion?

This is the first one. What's a well defined function, and what does the equation state in English?

thanks

Answer 1

To me, it is not a well - defined function since the f :R--> R but \(\sqrt{x+1}\) have to be positive.
Think of this, if x = -5\((-5\in R\)), then f(x) is undefined , right?

Answer 2

he beat me lol

Answer 3

f(x) is undefined because you can't sqrt a negative and get a real number?

Answer 4

A well-defined function is a function which is defined for all values of x
That is no matter what x is, you have f(x)
In this case, you don't have f(x) when x< -1

Answer 5

Why don't you have f(x) when x<-1? Is it because you have the square root of a negative number?

Answer 6

yes

Answer 7

If f: R-->C
then, it is well-defined

Answer 8

but in Real, you don't have \(\sqrt{negative~~numbers}\)

Answer 9

so for example:

f: [o,infinity) -> [0,infinity) : x |-> x^1/3

That would be a well defined function because from the stated domain, f(x) will be defined?

Answer 10

yes, it is well defined on R also, because it is \(\huge\sqrt[3]{x}\) And you have negative value for it, also
For example \(\huge \sqrt[3]{-8}=-2 \) because \((-2)^3 = -8 \)

Answer 11

for square root \(\sqrt{}\) , it need to have positive values,
for cube root , no need to have positive values

Answer 12

yes but the [0,infinity) being the domain, means that we don't have to take negative numbers into account anyway? So even if it was a square root, it would still be defined because negative isn't in the domain?

Answer 13

Yes,

Answer 14

okay wonderful thankyou. I think i understand it

Answer 15

:)