Question

Find the domain of the function $g\left(x\right)=\frac{\sqrt{x}}{2x^2+x-1}$.

Short answer

The domain of the function is $x\in \left[0,\infty \right)-\left\{\frac{1}{2}\right\}$.

Step-by-step solution

Step 1. By definition of domain of a function.

The given function is $g\left(x\right)=\frac{\sqrt{x}}{2x^2+x-1}$. Now, by definition of domain of function, the domain of the function is defined for all real values of $\left(x\right)$for which the function is defined. But $g\left(x\right)=\frac{\sqrt{x}}{2x^2+x-1}$is a rational function, therefore its denominator $2x^2+x-1$cannot be zero for the domain of the function to exist.

role="math" localid="1644589491038" $$\begin{gathered} 2x^2+x-1\ne 0 \\ \left(2x-1\right)\left(x+1\right)\ne 0 \\ x\ne \frac{1}{2},-\mathrm{1....}\left(\begin{array}{l}\text{forthedomain}\\ \text{of}g\left(x\right)\text{toexist}\end{array}\right)\mathrm{....}\left(1\right) \end{gathered}$$

Step 2. By definition of domain of square root function.

From the function $g\left(x\right)=\frac{\sqrt{x}}{2x^2+x-1}$, the domain is defined for all real numbers $\left(x\right)$for which the function $g\left(x\right)=\frac{\sqrt{x}}{2x^2+x-1}$is defined. Since the numerator of the function $g\left(x\right)=\frac{\sqrt{x}}{2x^2+x-1}$is the square root function, therefore the domain of function is defined for all real numbers except 3 for which the expression $\sqrt{x}\ge 0$.

role="math" localid="1644589657682" $$\begin{gathered} \text{So,}\sqrt{x}\ge 0\text{\hspace{0.33em}gives,} \\ x\ge 0....\left(2\right) \end{gathered}$$

Combining (1) and (2) the domain of the function is$x\in \left[0,\infty \right)-\left\{\frac{1}{2}\right\}$

Step 3. Description of steps.

The domain of the function is $x\in \left[0,\infty \right)-\left\{\frac{1}{2}\right\}$.