Domain and Range of Square Root Function #1

[Jason76]

Regarding

\(\displaystyle y=\sqrt x\)

y = function

y = range

\(\displaystyle y = \sqrt{x}\)

so the range is 0 and all positive numbers. These positive numbers and 0 can be found by taking the square root of x.

However, what about

\(\displaystyle y = \sqrt{x - 3}\) ?

Wouldn't the first 2 positive numbers (as well as 0) minus 3 = a negative number? A square root cannot be taken of a negative number.

 

[mmm4444bot]

Regarding

\(\displaystyle y=\sqrt x\)

y = function

y = range ← Do not write this

\(\displaystyle y = \sqrt{x}\)

so the range is 0 and all positive numbers. These positive numbers and 0 can be found by taking the square root of x.

Yes -- the range of sqrt(x) is 0 and all positive numbers, and the domain sqrt(x) is the same set of numbers as the range, but instead of saying "0 and all positive numbers" we say:

Domain: x ≥ 0

Range: y ≥ 0


Also, it's not correct to say that symbol y equals the range. It doesn't because symbol y is a variable. (These exercises don't usually require you to assign a symbol to name the set of numbers which is the range. You may simply call it "Range".)

However, what about


\(\displaystyle y = \sqrt{x-3}\) ?


Wouldn't the first 2 positive numbers (as well as 0) minus 3 = a negative number? A square root cannot be taken of a negative number.

Okay -- now you're talking about the Domain (the set of allowable x-values).

But what are you thinking, when you say "the first 2 positive numbers"? It seems like you're thinking about Integers, but x is not restricted to just Integers.

The correct way to determine the domain of sqrt(x-3) is to write an inequality that says the radicand must be non-negative.

x - 3 ≥ 0

Solve this inequality for x, and that will give you the domain.

Cheers :cool:

 

[soroban]

Hello, Jason76!

Regarding: \(\displaystyle y=\sqrt x\)

The domain is: .\(\displaystyle \color{blue}{x \,\ge\,0}\)

So the range is 0 and all positive numbers.
These positive numbers and 0 can be found by taking the square root of \(\displaystyle x.\)

However, what about: \(\displaystyle y = \sqrt{x - 3}\) ?

Wouldn't the first 2 positive numbers (as well as 0) minus 3 = a negative number?
A square root cannot be taken of a negative number.
Yes, but why are you concerned with integers?

You are right . . . we cannot have the square root of a negative number.
So \(\displaystyle x-3\) must be zero or greater.

Hence, we have: .\(\displaystyle x - 3 \:\ge\:0 \quad\Rightarrow\quad x \:\ge\:3\)

The domain is: \(\displaystyle x \,\ge\,3\)

And the range is: \(\displaystyle y \,\ge\,0\)

 

[srmichael]

Regarding

\(\displaystyle y=\sqrt x\)

y = function

y = range

\(\displaystyle y = \sqrt{x}\)

so the range is 0 and all positive numbers. These positive numbers and 0 can be found by taking the square root of x.

However, what about

\(\displaystyle y = \sqrt{x - 3}\) ?

Wouldn't the first 2 positive numbers (as well as 0) minus 3 = a negative number? A square root cannot be taken of a negative number.

In general, the domain of \(\displaystyle y = \sqrt{whatever}\) means we need "whatever" >= 0. Thus, if "whatever" = x, then x >= 0 is the domain as you noted. Now, if "whatever" = x - 3, then x - 3 >= 0, so, solving for x, the domain is x >= 3. Make sense?

P.S. If \(\displaystyle y = \sqrt{whatever}\) is in the denominator, then the domain changes to "whatever" > 0, since we can't have 0 in the denominator.

 

[mmm4444bot]

Jason -- I moved your next exercise into a new thread.