Answer
See below:
Work Step by Step
Simplify the provided inequality by adding 3 to both sides,
$4{{x}^{2}}+7x+3<0$
The boundary points of the given inequality will be calculated by equating $f\left( x \right)$ to 0 as $4{{x}^{2}}+7x+3=0$.
Thus,
$\begin{align}
& 4{{x}^{2}}+4x+3x+3=0 \\
& 4x\left( x+1 \right)+3\left( x+1 \right)=0 \\
& \left( 4x+3 \right)\left( x+1 \right)=0
\end{align}$
Hence,
$x=-1,x=-\frac{3}{4}$
These values of x are the boundary points so locate these point on the number line.
From the above number line, the boundary points divide the number line into three parts as,
$\left( -\infty ,-1 \right)\text{,}\left( -1,-\frac{3}{4} \right)\text{ and }\left( -\frac{3}{4},\infty \right)$
Now, one test value within each interval is chosen and f is evaluated at that number.
As can be observed, for both intervals $\left( -\infty ,-1 \right)\text{ and }\left( -\frac{3}{4},\infty \right)$ , the function is positive.
And, for the interval $\left( -1,-\frac{3}{4} \right)$ the function is negative.
Also, the points $-1\text{ and }-\frac{3}{4}$ are not included in the solution because the inequality is strict, that is, it does not include the equality sign.
Hence, the required interval is $\left( -1,-\frac{3}{4} \right)$.
