Question

How do you find the domain of `f(x)=x^2+4`?

Answer 1

The domain of a function is the set of the values in which you can calculate the function itself.

Think of a function as a robot, which takes as an input a number, and gives you back, as an output, another number.

The problem of finding the domain of a functions is due to the fact that not every function "accepts" every real number as an input.

There are three main restrictions:

1. If you have a rational function, you must exclude from the domain the values for which the denominator is zero, since you cannot divide by zero. So, for example, if you consider the function `f(x)=\frac{1}{x-2}`, you see that you can evaluate it for every real value `x`, as long as it is not 2: in that case, you would have `f(2)=\frac{1}{2-2}`, an obviously invald operation. So, we say that the domain of `f` is the whole real number set, deprived of the element "2".

2. The second restriction is about (even) roots: I assume you know that you can't calculate the square root of a (strictly) negative number, since such a number does not exist in the real set. So, the domain of the function `f(x)=\sqrt{x}` is the set `{x \in \mathbb{R} | x\ge0}`.

3. The same thing goes for the logarithms, as they only "accepts" positive numbers as an input.

So, every time you have to find the domain of a function, you simply need to make sure that all the objects involved are well-defined, or, if you want to read it in terms of robots, you have to make sure that you won't give the robot an input he will not accept.

Now, let's find the domain of your function. It is not a rational function, and there are no roots of logaritmhs. We have no requests to satisfy, and thus the domain of the function is the set of all the real numbers, `\mathbb{R}`.

Your function is indeed a polynomial, and it is a general rule that a polynomial is always well-defined over all the real axis. After all, try to see it this way: your function `f` takes a number, squares it, and the adds 4 to the result. For which number you shouldn't be able to do so?:)