Domain and Range of the function $\frac{1}{2-\sqrt{x}}$

Question

$$\frac{1}{2-\sqrt{x}}$$

I guess the domain should be $x \geq 0$ since $x \geq 0$, $2 - \sqrt{x} \leq 2$.

But how do I evaluate the range?

Answer 1

For the domain, $x\geq0$ (due to the square root) AND $2-\sqrt{x}\neq0$ (can you tell why?)

For the range, note that if $2-\sqrt{x}\to0^+,f(x)\to+\infty$ and $2-\sqrt{x}\to0^-,f(x)\to-\infty$. Also, since $x\geq0$, $f(x)_{min}$ will exist. Consider graphing the function for a complete picture.

Answer 2

Domain $D$:$ [0,4)\cup(4,\infty).$

Set $y:√x$ , $y \in D_y= [0,2) \cup (2,\infty).$

$f(y):= \dfrac{1}{2-y}$.

1) Let $y \in [0,2).$

Then: Image $f = [1/2,\infty )$

2) Let $y \in (2,\infty)$.

Then: Image $f= (-\infty, 0).$

Hence : Image $f = (-\infty,0)\cup [1/2,\infty).$