Question

The domain of the function $\sqrt{x^2-5x+6}+\sqrt{2x+8-x^2}$ is

• A. $\left[2,3\right]$
• B. $\left[-2,4\right]$
• C. $\left[-2,2\right]\cup \left[3,4\right]$
• D. $\left[-2,1\right]\cup \left[2,4\right]$

Answer 1

The correct option is C

$\left[-2,2\right]\cup \left[3,4\right]$

Explanation for the correct option(s)

Finding the domain of the function $f\left(x\right)$

Given function $f\left(x\right)=\sqrt{\left(x^2-5x+6\right)}+\sqrt{\left(2x+8-x^2\right)}$ has two domains involved in it.

Both are square root functions and it is defined to be positive always. Hence the functions must be $\ge 0$.

Function 1: $x^2-5x+6$

$$\begin{gathered} \Rightarrow x^2-5x+6\ge 0 \\ \Rightarrow x^2-2x-3x+6\ge 0\left[Rewrite-5xas-2x-3x\right] \\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)\ge 0 \\ \Rightarrow \left(x-3\right)\left(x-2\right)\ge 0 \\ \therefore x(-\infty ,2]\cup [3,\infty ) \end{gathered}$$

Function 2:

$$\begin{gathered} \Rightarrow -x^2+2x+8\ge 0 \\ \Rightarrow -x^2+4x-2x+8\ge 0\left[\mathrm{Rewrite}2x\mathrm{as}4x-2x\right] \\ \Rightarrow x\left(-x+4\right)+2\left(-x+4\right)\ge 0 \\ \Rightarrow \left(-x+4\right)\left(x+2\right)\ge 0 \\ \Rightarrow x\ge -2o\mathrm{r}x\le 4 \\ \therefore x\left[-2,4\right] \end{gathered}$$

The final domain is determined basis the intersection point of the above functions. Hence, the domain of the given function is $\left[-2,2\right]\cup \left[3,4\right]$

Therefore, the correct answer is Option (C).