Determine the domain

[spacewater]

problem
determine the domains of (a) \(\displaystyle f\), (b)\(\displaystyle g\), and (c) \(\displaystyle f \cdot g\)
\(\displaystyle f(x) = \sqrt{x+4} , g(x) = x^2\)

domain of f - any real number except -4 since the answer will be 0

domain of g - shouldn't the answer be any real number except 0? I don't understand why 0 is included in domain



Can someone elaborate this problem to me please?

 

[galactus]

problem
determine the domains of (a) \(\displaystyle f\), (b)\(\displaystyle g\), and (c) \(\displaystyle f \cdot g\)
\(\displaystyle f(x) = \sqrt{x+4} , g(x) = x^2\)

\(\displaystyle f=\sqrt{x+4}\), domain of f - any real number except -4 since the answer will be 0

The domain are the acceptable values that give a real result. 0 is cool. \(\displaystyle \sqrt{0}=0\). That's perfectly fine.

What you do not want are negatives inside the radical. Such as \(\displaystyle \sqrt{-5+4}=\sqrt{-1}=i\). No good. -5 is NOT in the domain.

Therefore, what is NOT in the domain is \(\displaystyle (-\infty, -4)\)

Everything from negative infinity up to but not including -4 will result in a negative inside the radical. So, the domain is everything but that.

Domain would be \(\displaystyle [-4,\infty)\) because those are the parameters that give us real solutions in the range. The bracket means we include -4. A parenthesis would be

everything up to, but not including, 4. See now?.

Since we can have no negative results from a square root, think of it this way: \(\displaystyle \sqrt{x+4}\geq 0\).

Square both sides. Yes, you can do that. \(\displaystyle x+4\geq 0\)

\(\displaystyle x\geq -4\). There's the domain. All values of x greater than or equal to -4.

Think of the domain as what goes in and the range as what comes out. The range would be \(\displaystyle [0,\infty)\)

You seem to think that 0 is not allowed. It is. The domain of \(\displaystyle x^{2}\) would be \(\displaystyle (-\infty, \infty)\) because we can plug in anything in the reals for x and get a real result. \(\displaystyle 0^{2}=0\), \(\displaystyle (-1000)^{2}=1000000\) and so on.