Find the domain of the functions $\sqrt {5-x} $ and $3x^2+\frac {6 } {x } -8$ Given the function $h(x)=\sqrt {5-x} $ and the function $f(x)=3x^2+\frac {6 } {x }  -8$ how would I find the domain without graphing?
 A: Hints:
$\bullet$ The real function $\,\sqrt x\,$ is defined only for $\,x\ge 0\;$ ;
$\bullet$ Rational functions (i.e. polynomial function divided by polynomial function) are defined everywhere except at the denominator's zeros.
A: Start with all real numbers and throw away any value that causes an error (or keep the ones that do not cause an error, whichever is easier to consider).
For $\sqrt{5 - x}$, an error occurs when you try to take the square root of a negative number. The error-causing values are therefore all real numbers $x$ such that
$$
5 - x < 0.
$$
For $3x^2 + \frac{6}{x} - 8$, an error occurs when you try to divide by zero.
A: Suppose you have two functions $f$ and $G$.
Let $g$ have domain $D_g$.
Take the relational composite $g \circ f$.
Then $g \circ f$ has domain $f^{-1}[D_g]$.
So in the case of $h(x) = \sqrt{5-x}$, $h = g\circ f$, where $f(x) = 5-x$ and $g(x)=\sqrt x$.
Assuming you're dealing with real square roots, $D_g = {\uparrow}0$, so you need to solve the inequality $5-x>0$.
