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Chapter 2: Problem 17

Solve each polynomial inequality and graph the solution set on a real number line. Express each solution set in interval notation. $$5 x \leq 2-3 x^{2}$$

Short Answer

Expert verified
The solution to the inequality is \( (-\infty,-1]\cup[\frac{2}{3},+\infty) \).

Step by step solution

01

Rearrange the inequality

The first step is to rearrange the inequality so that all terms are on one side and zero is on the other. This can be achieved by adding \(3x^2\) and subtracting \(5x\) from both sides of the inequality \(5x \leq 2 - 3x^2\). This results in \(3x^2 + 5x - 2 \geq 0\).
02

Solve the quadratic equation

Next, solve the equation \(3x^2 + 5x - 2 = 0\) to find the critical points where the inequality potentially may change sign. Quadratic equation can be solved by factorization, completing the square or quadratic formula. The roots of the equation are \(x = \frac{-5 \pm \sqrt {5^2 - 4*3*(-2)}}{2*3}\). Calculating gives the roots \(x = -1\) and \(x = \frac {2}{3}\)
03

Determine the intervals and test the inequality

Having obtained the roots of the equation, the real number line is divided into three intervals: \(-\infty, -1\), \(-1, \frac {2}{3}\), and \(\frac {2}{3}, +\infty\). Choose test points from each interval, say \(-2\), \(0\) and \(1\), and substitute them into the inequality \(3x^2 + 5x - 2 \geq 0\). This gives, for \(x = -2, 3*(-2)^2 + 5*(-2) - 2 \geq 0\) which is true. For \(x = 0, 3*(0)^2 + 5*0 - 2 \geq 0\) which is false. And for \(x = 1, 3*(1)^2 + 5*1 - 2 \geq 0\) which is true. The tested points satisfy the inequality in intervals \(-\infty, -1\) and \(\frac {2}{3}, +\infty\).
04

Writing the solution in interval notation

The intervals which satisfy the inequality are represented in interval notation as \((-\infty,-1] \cup [\frac{2}{3}, +\infty)\). The square brackets indicate that the given point is included in the solution set.

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Most popular questions from this chapter

Use everyday language to describe the behavior of a graph near its vertical asymptote if \(f(x) \rightarrow \infty\) as \(x \rightarrow-2^{-}\) and \(f(x) \rightarrow-\infty\) as \(x \rightarrow-2^{+}\).

Will help you prepare for the material covered in the next section. Solve: \(x^{3}+x^{2}=4 x+4\)

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. \((x+3)(x-1) \geq 0\) and \(\frac{x+3}{x-1} \geq 0\) have the same solution set.

Find the horizontal asymptote, if there is one, of the graph of rational function. $$h(x)=\frac{12 x^{3}}{3 x^{2}+1}$$

Use transformations of \(f(x)=\frac{1}{x}\) or \(f(x)=\frac{1}{x^{2}}\) to graph each rational function. $$g(x)=\frac{1}{x+2}-2$$
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