What is the domain of $f(x)=\sqrt{2x-8}$?

Question

Is the domain of $f(x)$ simply any number, but in this Khan Academy video, it said the domain is $x \geqslant 4$.

Answer 1

If function is defined on real numbers , then the square root expression must be non negative , because $\sqrt -a$ for positive $a$ is not defined in real numbers

So expression under radical is $2x-8$

Therefore $$2x-8\ge 0$$ it gives $$ x\ge 4$$ so the domain is $$x\in [4,\infty)$$

Answer 2

$2x-8 \geqslant 0$ implies $x\geqslant 4$.

Answer 3

If $x<4$, then $2x-8<0$ and $\sqrt{2x-8}$ doesn't exist. (At least, it has no generic definition and it won't be a real number even if you build an extension of square root on negative, or even complex, numbers.)

Thus you can only compute $f(x)$ when $x\geq 4$, that's why the domain of $f$ is $[4,+\infty)$

Answer 4

Since $f$ is a real-valued function of a real variable, unless otherwise specified, its domain is understood to be the set of all real numbers $x$ such that $f(x)$ is real-valued. For a square root function, this means the radicand, the term inside the radical, must be nonnegative since no negative number has a real-valued square root. Hence, for $x$ to be in the domain of $f$, \begin{align*} 2x - 8 & \geq 0\\ 2x & \geq 8\\ x & \geq 4 \end{align*}
Thus, the domain of $f$ is $\text{Dom}_f = [4, \infty)$.