## Precalculus (6th Edition) Blitzer

Convert the function in the form of $f\left( x \right)\ge 0$ or $f\left( x \right)\le 0$. \begin{align} & 5x\le 2-3{{x}^{2}} \\ & 5x-2+3{{x}^{2}}\le 2-3{{x}^{2}}-2+3{{x}^{2}} \\ & 3{{x}^{2}}+5x-2\le 0 \end{align} The boundary points of the given inequality will be calculated by equating $f\left( x \right)$ to 0 as $3{{x}^{2}}+5x-2=0$. Now, use $\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Here, $a=3\text{,b}=5,\text{c}=-2$ So, \begin{align} & x=\frac{-\left( 5 \right)\pm \sqrt{{{5}^{2}}-4\times 3\times \left( -2 \right)}}{2\times 3} \\ & x=\frac{-5\pm \sqrt{25+24}}{6} \\ & x=\frac{-5\pm 7}{6} \\ & x=-2,\frac{1}{3} \end{align} Hence, $x=-2 \text{ and }\frac{1}{3}.$ These values of x are the boundary points so locate these points on the number line as, Now as can be seen from the above number line, $x=\frac{1}{3}$ and $x=-2$ divides the number line in three intervals as, $\left( -\infty ,-2 \right],\left[ -2,\frac{1}{3} \right],\left[ \frac{1}{3},\infty \right)$ If $f\left( x \right)\ge 0$ or $f\left( x \right)\le 0$ then x includes the boundary point in its interval, so use big brackets for including points. Now, test the condition for $\left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0$ in interval $\left( -\infty ,-2 \right]$. Take test point $-8$ , $\begin{matrix} \left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0 \\ \left( -8-\frac{1}{3} \right)\left( -8+2 \right)\overset{?}{\mathop{\le }}\,0 \\ 50\overset{?}{\mathop{\le }}\,0 \\ \end{matrix}$ Thus, the condition is unsatisfied. Now, test the condition for $\left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0$ in interval $\left[ -2,\frac{1}{3} \right]$. Take test point $0$ as, $\begin{matrix} \left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0 \\ \left( 0-\frac{1}{3} \right)\left( 0+2 \right)\overset{?}{\mathop{\le }}\,0 \\ -\frac{2}{3}\overset{?}{\mathop{\le }}\,0 \\ \end{matrix}$ Thus, the condition is satisfied. Now, test the condition for $\left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0$ in interval $\left[ \frac{1}{3},\infty \right)$. Take test point $5$ , \begin{align} & \left( x-\frac{1}{3} \right)\left( x+2 \right)\le 0 \\ & \left( 7-\frac{1}{3} \right)\left( 7+2 \right)\overset{?}{\mathop{\le }}\,0 \\ & 60\overset{?}{\mathop{\le }}\,0 \end{align} Thus, the condition is unsatisfied. Therefore, the interval satisfying the given inequality is $\left[ -2,\frac{1}{3} \right]$. Thus, the solution set in the interval notation will be $\left[ -2,\frac{1}{3} \right]$.