Question

The domain of the function $f\left(x\right)=\sqrt{\left(x^{14}-x^{11}+x^6-x^3+x^2+1\right)}$ is

• A. $\left(-\infty ,\infty \right)$
• B. $[0,\infty )$
• C. $(-\infty ,0]$
• D. $\frac{R}{\left[0,1\right]}$

Answer 1

The correct option is A

$\left(-\infty ,\infty \right)$

Explanation for correct Option:

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

Given function $f\left(x\right)=\sqrt{\left(x^{14}-x^{11}+x^6-x^3+x^2+1\right)}$ is a square root function and can be true if $f\left(x\right)\ge 0$.

Rewriting the above function,

$$\begin{gathered} \Rightarrow \left(x^{14}-x^{11}+x^6-x^3+x^2+1\right)\ge 0 \\ \Rightarrow \left(x^{14}+x^6+x^2\right)-\left(x^{11}+x^3\right)+1\ge 0 \end{gathered}$$

Breaking the function further,

For positive value of $x$, the above function will always be positive.

For negative value of $x$,

$\left(x^{14}+x^6+x^2\right)$ will be positive as all values of $x$ in this function is powered to even numbers.

$\left(x^{11}+x^3\right)$ will be negative as all values of $x$ in this function is powered to odd numbers.

However, applying this on the overall function would mean

$$\begin{gathered} \Rightarrow \left(x^{14}+x^6+x^2\right)-\left(x^{11}+x^3\right)+1\ge 0 \\ \Rightarrow \left(+\right)-\left(-\right)+1\ge 0 \\ \Rightarrow \left(+\right)+\left(+\right)+1\ge 0 \end{gathered}$$

Hence, $x$ can be of any value and the domain of the function is $\left(-\infty ,\infty \right)$.

Hence, the correct answer is Option (A).