Domain of Square Root Function

• Sep 30th 2008, 09:29 PM
magentarita
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}
• Sep 30th 2008, 10:02 PM
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.
• Sep 30th 2008, 11:43 PM
mr fantastic
Quote:

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 $2x + 3 \geq 0$.

(2) Solve $2x - x^2 \geq 0$.
• Oct 1st 2008, 05:24 PM
magentarita
yes but......
Quote:

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

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

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

Yes, but why? I am looking for the why of things not just how to find the answer.
• Oct 1st 2008, 06:36 PM
Krizalid
Well, it's not that hard to find the "why."

Let's use our brain: having $f(x)=\sqrt x.$ For what values of $x,$ $\sqrt x$ is well defined? Yes, we require that $x\ge0,$ 'cause $x$ 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 $x\ge0.$

Now, by applyin' this to your problems, you'll see this does make sense.
• Oct 2nd 2008, 04:01 AM
magentarita
ok
Quote:

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

Let's use our brain: having $f(x)=\sqrt x.$ For what values of $x,$ $\sqrt x$ is well defined? Yes, we require that $x\ge0,$ 'cause $x$ 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 $x\ge0.$

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

ok...now I get it.