\begin{align*}
x^{2}  x  12 & < 0 \\
(x  4)(x + 3) & < 0
\end{align*}
Critical values 
 \(x=3\) 
 \(x=4\) 

\(x  4\) 
\(\) 
\(\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(x + 3\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(+\) 
\(+\) 
\(f(x)= (x  4)(x + 3)\) 
\(+\) 
\(\text{0}\) 
\(\) 
\(\text{0}\) 
\(+\) 
From the table we see that \(f(x)\) is less than \(\text{0}\)
when \(3 < x < 4\)
We represent this on a number line:
\begin{align*}
3x^{2} + x  4 & > 0 \\
(3x + 4)(x  1) & > 0
\end{align*}
Critical values 
 \(x=\frac{4}{3}\) 
 \(x=1\) 

\(x  1\) 
\(\) 
\(\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(3x + 4\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(+\) 
\(+\) 
\(f(x)= (3x + 4)(x  1)\) 
\(+\) 
\(\text{0}\) 
\(\) 
\(\text{0}\) 
\(+\) 
From the table we see that \(f(x)\) is greater than
\(\text{0}\) when \(x < \frac{4}{3}\) or when \(x
> 1\)
We represent this on a number line:
\[y^{2} + y + 2 < 0\]
There are no real solutions.
The graph lies above the \(x\)axis and does not cut the
\(x\)axis so the function is never negative. There are
no values of \(y\) that will solve this inequality.
\[(3  t)(1 + t) > 0\]
Critical values 
 \(t = 1\) 
 \(t = 3\) 

\(3  t\) 
\(+\) 
\(+\) 
\(+\) 
\(\text{0}\) 
\(\) 
\(1 + t\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(+\) 
\(+\) 
\(f(x)= (3  t)(1 + t)\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(\text{0}\) 
\(\) 
From the table we see that \(f(x)\) is greater than
\(\text{0}\) when \(1 < t < 3\)
We represent this on a number line:
\[s^{2}  4s + 6 > 0 \\\]
Use the quadratic formula to find critical values:
\begin{align*}
s & = \frac{(4) \pm \sqrt{(4)^2  4(1)(6)}}{2(1)} \\
& = \frac{4 \pm \sqrt{16 24}}{2 } \\
& = \frac{4 \pm \sqrt{8}}{2}
\end{align*}
Therefore there are no real roots and the graph does not cut
the \(x\)axis. The graph lies above the \(x\)axis and
so this inequality is true for all real values of \(s\).
Use the quadratic formula to find critical values:
\begin{align*}
x & = \frac{(1) \pm \sqrt{(1)^2  4(7)(8)}}{2(7)} \\
& = \frac{1 \pm \sqrt{1 224}}{14} \\
& = \frac{1 \pm \sqrt{223}}{14}
\end{align*}
Therefore there are no real roots and the graph does not cut
the \(x\)axis. The graph lies above the \(x\)axis and
so this inequality is true for all real values of \(x\).
\begin{align*}
x & \geq 4x^2 \\
4x^2 + x & \geq 0 \\
x(4x + 1) & \geq 0
\end{align*}
Critical values 
 \(x = \frac{1}{4}\) 
 \(x = 0\) 

\(x\) 
\(\) 
\(\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(4x + 1\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(+\) 
\(+\) 
\(f(x)= x(4x + 1)\) 
\(+\) 
\(\text{0}\) 
\(\) 
\(\text{0}\) 
\(+\) 
From the table we see that \(f(x)\) is greater than
\(\text{0}\) when \(x \leq \frac{1}{4}\) or \(x \geq
0\).
We can represent this on a number line:
\[2x^{2} + x + 6 \le 0\]
There are no real roots and the graph does not cut the
\(x\)axis. The graph lies above the \(x\)axis and so
this inequality is never true.
\(\dfrac{x}{x3}<2\), \(x\ne 3\)
We first solve the equation:
\begin{align*}
\dfrac{x}{x3} & < 2 \\
\dfrac{x}{x3}  2 & < 0 \\
\dfrac{x  2(x3)}{x3}& < 0 \\
\dfrac{x  2x + 6}{x3}& < 0 \\
\dfrac{x + 6}{x3}& < 0 \\
\dfrac{(x  6)}{x3}& < 0 \\
\dfrac{x  6}{x3}& > 0 \\
x & = 6
\end{align*}
Critical values 
 \(x = 3\) 
 \(x = 6\) 

\(x  3\) 
\(\) 
undef 
\(+\) 
\(+\) 
\(+\) 
\(x  6\) 
\(\) 
\(\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(f(x)= x  6\) 
\(+\) 
undef 
\(\) 
\(\text{0}\) 
\(+\) 
From the table we see that \(f(x) > 0\) when \(x < 3
\text{ or } x > 6\) with \(x \ne 3\).
We can represent this on a number line:
\(\dfrac{x^2 + 4}{x  7} \geq 0\), \(x \ne 7\)
We first simplify the fraction:
\[\dfrac{(x + 2)(x  2)}{x  7} \geq 0\]
Critical values 
 \(x = 2\) 
 \(x = 2\) 
 \(x = 7\) 

\(x + 2\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(+\) 
\(+\) 
\(+\) 
\(+\) 
\(x  2\) 
\(\) 
\(\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(+\) 
\(+\) 
\(x  7\) 
\(\) 
\(\) 
\(\) 
\(\) 
\(\) 
undef 
\(+\) 
\(f(x)= \dfrac{(x + 2)(x  2)}{x  7}\) 
\(\) 
\(\text{0}\) 
\(+\) 
\(\text{0}\) 
\(\) 
undef 
\(+\) 
From the table we see that \(f(x)\) is greater than
\(\text{0}\) when \(2 \leq x \leq 2\) and \(x > 7\)
with \(x \ne 7\).
We can represent this on a number line:
\(\dfrac{x + 2}{x}  1 \geq 0\), \(x \ne 0\)
We first simplify the equation:
\begin{align*}
\frac{x + 2}{x}  1 & \ge 0 \\
\frac{x + 2  x}{x}& \ge 0 \\
\frac{2}{x} & \ge 0 \\
\text{Therefore } x & > 0
\end{align*}
The solution is \(x > 0\) with \(x \ne 0\).
We can represent this on a number line: