Thread: Domain of Square Root Function

1. Domain of Square Root Function

How do you find the domain and range of square root functions?

(1) Find the domain of y = sqrt{2x + 3}

(2) Find the domain of f(x) = sqrt{2x - x^2}

2. Well since we're dealing with real numbers, we aren't allowed to take the square-root of a negative number. So find out for what x-values the expression inside the square-root is less than 0.

3. Originally Posted by Jameson
Well since we're dealing with real numbers, we aren't allowed to take the square-root of a negative number. So find out for what x-values the expression inside the square-root is less than 0.
Just to expand a bit, the domains are given by the solution to the following inequations:

(1) Solve $\displaystyle 2x + 3 \geq 0$.

(2) Solve $\displaystyle 2x - x^2 \geq 0$.

4. yes but......

Originally Posted by mr fantastic
Just to expand a bit, the domains are given by the solution to the following inequations:

(1) Solve $\displaystyle 2x + 3 \geq 0$.

(2) Solve $\displaystyle 2x - x^2 \geq 0$.
Yes, but why? I am looking for the why of things not just how to find the answer.

5. Well, it's not that hard to find the "why."

Let's use our brain: having $\displaystyle f(x)=\sqrt x.$ For what values of $\displaystyle x,$ $\displaystyle \sqrt x$ is well defined? Yes, we require that $\displaystyle x\ge0,$ 'cause $\displaystyle x$ can be zero, or positive, but it can't be negative, since square roots of negative numbers don't belong to the Real numbers, that's why we require $\displaystyle x\ge0.$

Now, by applyin' this to your problems, you'll see this does make sense.

6. ok

Originally Posted by Krizalid
Well, it's not that hard to find the "why."

Let's use our brain: having $\displaystyle f(x)=\sqrt x.$ For what values of $\displaystyle x,$ $\displaystyle \sqrt x$ is well defined? Yes, we require that $\displaystyle x\ge0,$ 'cause $\displaystyle x$ can be zero, or positive, but it can't be negative, since square roots of negative numbers don't belong to the Real numbers, that's why we require $\displaystyle x\ge0.$

Now, by applyin' this to your problems, you'll see this does make sense.
ok...now I get it.