Answer
See below:
Work Step by Step
Simplify the provided inequality by subtracting 1 from both sides,
$6{{x}^{2}}+x-1>0$
The boundary points of the given inequality will be calculated by equating $f\left( x \right)$ to 0 as $6{{x}^{2}}+x-1>0$
Thus,
$\begin{align}
& 6{{x}^{2}}+3x-2x-1=0 \\
& 3x\left( 2x+1 \right)-1\left( 2x+1 \right)=0 \\
& \left( 3x-1 \right)\left( 2x+1 \right)=0
\end{align}$
Hence,
$x=\frac{1}{3},x=-\frac{1}{2}$
From the above number line, the boundary points divide the number line into three parts as,
$\left( -\infty ,-\frac{1}{2} \right)\text{,}\left( -\frac{1}{2},\frac{1}{3} \right)\text{ and }\left( \frac{1}{3},\infty \right)$.
Now, one test value within each interval is chosen and f is evaluated at that number.
As can be observed, for both intervals $\left( -\infty ,-\frac{1}{2} \right)\text{ and }\left( \frac{1}{3},\infty \right)$ , the function is positive.
And, for interval $\left( -\frac{1}{2},\frac{1}{3} \right)$ , the function is negative.
And, the points $-\frac{1}{2}\text{ and }\frac{1}{3}$ are not included in the solution because inequality is strict, that is, it does not include the equality sign.
Thus, the solution set in the interval notation will be $\left( -\infty ,-\frac{1}{2}, \right)\cup \left( \frac{1}{3},\infty \right)$.
