Answer
See below:
Work Step by Step
Consider the inequality,
$9{{x}^{2}}+3x-2\ge 0$
The boundary points of the given inequality will be calculated by equating $f\left( x \right)$ to 0 as $9{{x}^{2}}+3x-2=0$.
Thus,
$\begin{align}
& 9{{x}^{2}}+6x-3x-2=0 \\
& 3x\left( 3x+2 \right)-1\left( 3x+2 \right)=0 \\
& \left( 3x-1 \right)\left( 3x+2 \right)=0
\end{align}$
Hence,
$x=\frac{1}{3},x=-\frac{2}{3}$
As can be seen from the above number, the boundary points divide the number line into three parts,
$\left( -\infty ,-\frac{2}{3} \right)\text{,}\left( -\frac{2}{3},\frac{1}{3} \right)\text{ and }\left( \frac{1}{3},\infty \right)$.
Now, one test value within each interval is chosen and f is evaluated at that number.
So, for both the interval $\left( -\infty ,-\frac{2}{3} \right)\text{ and }\left( \frac{1}{3},\infty \right)$ the function is positive.
And, for interval $\left( -\frac{2}{3},\frac{1}{3} \right)$ the function is negative.
Also, the points $\frac{-2}{3}\text{ and }\frac{1}{3}$ are included in the solution because the inequality is not strict, that is, it includes the equality sign.
Hence, the required interval is $\left( -\infty ,-\frac{2}{3} \right]\text{ and }\left[ \frac{1}{3},\infty \right)$.
