Graphs of irrational functions

The inside of $\sqrt{\ }$ must be at least $0$, so for $y=\sqrt{2x-10}$, $2x-10\ge 0$ gives the domain $x\ge 5$. For $y=\sqrt{4-x}$, $4-x\ge 0$ gives $x\le 4$: the inequality flips.

Since $\sqrt{x}\ge 0$, the range of $y=a\sqrt{x}+q$ depends on the sign of $a$.

For example, the range of $y=3\sqrt{x}+4$ is $y\ge 4$, and the range of $y=-\sqrt{x}+2$ is $y\le 2$.

Examples

1. 01Which one has all real numbers as its domain?$\displaystyle$A$\displaystyle y=\sqrt{x^{2}+3}$B$\displaystyle y=\sqrt{x-2}$C$\displaystyle y=\sqrt{x}$D$\displaystyle y=\sqrt{x^{2}-3}$AnswerHide

   Answer$\displaystyle y=\sqrt{x^{2}+3}$

   Steps

   1. Look for an inside that is never negative
   2. $\displaystyle x^{2}+3$$\displaystyle \geq 3 > 0$
   3. Meanwhile $x^2-3$ goes negative for x near 0
2. 02What goes in □?$\displaystyle y=\sqrt{2x-10} \;\Rightarrow\; x \geq \square$AnswerHide

   Answer$\displaystyle 5$

   Steps

   1. The inside of the root must be ≥ 0
   2. $\displaystyle 2x-10$$\displaystyle \geq 0$
   3. Divide by 2
   4. $\displaystyle x$$\displaystyle \geq 5$
3. 03What is the domain?$\displaystyle y=\sqrt{4-x}$AnswerHide

   Answer$\displaystyle x \leq 4$

   Steps

   1. The inside of the root must be ≥ 0
   2. $\displaystyle 4-x$$\displaystyle \geq 0$
   3. Move x to the right-hand side
   4. $\displaystyle 4$$\displaystyle \geq x$
   5. Swapping the two sides swaps the direction too
   6. $\displaystyle x$$\displaystyle \leq 4$
4. 04What is the range?$\displaystyle y=3\sqrt{x}+4$AnswerHide

   Answer$\displaystyle y \geq 4$
5. 05What goes in □?$\displaystyle y=\sqrt{3x-15} \;\Rightarrow\; x \geq \square$AnswerHide

   Answer$\displaystyle 5$

   Steps

   1. The inside of the root must be ≥ 0
   2. $\displaystyle 3x-15$$\displaystyle \geq 0$
   3. Divide by 3
   4. $\displaystyle x$$\displaystyle \geq 5$
6. 06What is the range?$\displaystyle y=-\sqrt{x}+2$AnswerHide

   Answer$\displaystyle y \leq 2$

   Steps

   1. $\sqrt{x}$ is ≥ 0, so with a minus in front it is ≤ 0
   2. $\displaystyle -\sqrt{x}$$\displaystyle \leq 0$
   3. Then 2 is added
   4. $\displaystyle y$$\displaystyle \leq 2$

FAQ

Q1Why isn't the domain of $y=\sqrt{4-x}$ $x\ge 4$?AnswerHide

A

Solving $4-x\ge 0$ gives $-x\ge -4$. Multiplying both sides by $-1$ flips the inequality, so $x\le 4$.

Q2Can a square root function be defined for all real numbers?AnswerHide

A

Yes. In $y=\sqrt{x^2+3}$ the inside is always at least $3$, so it is defined for every $x$.