Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.7 - Polynomial and Rational Inequalities - Exercise Set - Page 412: 16


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Work Step by Step

Consider the provided inequality $3{{x}^{2}}+16x+5<0$ The boundary points of the given inequality will be calculated by equating $f\left( x \right)$ to 0 as $3{{x}^{2}}+16x+5=0$ Thus, $\begin{align} & 3{{x}^{2}}+x+15x+5=0 \\ & x\left( 3x+1 \right)+5\left( 3x+1 \right)=0 \\ & \left( x+5 \right)\left( 3x+1 \right)=0 \end{align}$ Hence, $x=-5,x=-\frac{1}{3}$ 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, $\left( -\infty ,-5 \right)\text{,}\left( -5,-\frac{1}{3} \right)\text{ and }\left( -\frac{1}{3},\infty \right)$. Now, one test value within each interval is chosen and f is evaluated at that number. As can be observed, for both intervals $\left( -\infty ,-5 \right)\text{ and }\left( -\frac{1}{3},\infty \right)$ , the function is positive. And, for interval $\left( -5,-\frac{1}{3} \right)$ , the function is negative. Also, the points $-5\text{ and }-\frac{1}{3}$ are not included in the solution because the function is not equal to 0. Hence, the required interval is $\left( -5,-\frac{1}{3} \right)$.
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