Question

If $f\left(x\right)=\sqrt{\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-4.5\right)}$, then $D\_f\left(x\right)$ is

• A. $(-\infty ,4]$
• B. $[3,\infty )$
• C. $(-\infty ,1]\cup [3,4]\cup [4.5,\infty )$
• D. $[3,4]\cup [4.5,\infty )$

Answer 1

The correct option is C

$(-\infty ,1]\cup [3,4]\cup [4.5,\infty )$

Explanation for the correct option.

Find the domain of $\left\{x\right\}$:

Given,

$f\left(x\right)=\sqrt{\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-4.5\right)}$.

$f\left(x\right)$ is define if $\left(x-1\right)\left(x-3\right)\left(x-4\right)\left(x-4.5\right)\ge 0\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

This inequality can be solved by the method of intervals.

If $x>4.5$, then the product of all the factors are positive. Equation $\left(1\right)$ is satisfied.

If $4<x<4.5$, then $\left(x-1\right),\left(x-3\right),\left(x-4\right)$ are positive but $(x-4.5)$ is negative, then the product of all the factors are not positive. Equation $\left(1\right)$ is not satisfied.

If $3<x<4$, then $\left(x-1\right),\left(x-3\right)$ are positive but others are negative, then the product of all the factors are positive. Equation $\left(1\right)$ is satisfied.

Therefore, $x\in [3,4]\cup [4.5,\infty )\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Again if $x<1$, then the product of all the factors are positive. Equation $\left(1\right)$ is satisfied.

Therefore, $x\in (-\infty ,1]\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

From relation $\left(2\right)$ and $\left(3\right)$,

$\Rightarrow x\in (-\infty ,1]\cup [3,4]\cup [4.5,\infty )$

Hence, the correct option is C.