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Chapter 2: Problem 40

Solve each polynomial inequality and graph the solution set on a real number line. Express each solution set in interval notation. $$x^{3}-x^{2}+9 x-9>0$$

Short Answer

Expert verified
The solution set is \((-\infty, 1) \cup (1,\infty)\).

Step by step solution

01

Factor the Polynomial

Factor the given polynomial. The standard approach to factor \(x^3-x^2+9x-9\) is to look for noticeable factors in the polynomial, in this case \(x-1\) and \(x^2+9\) are the factors. So you have \((x-1)(x^2+9)\).
02

Find the Critical Points

Set each factor to zero to find the critical points of the inequality. For \(x-1=0\), you have \(x=1\) as the critical point. The factor \(x^2+9\) doesn't yield any real solutions. So, in this case the only critical point is \(x=1\).
03

Test the Intervals

Choose test points in the intervals \(x<1\) and \(x>1\) and plug them into the factored form of the inequality. Let's choose \(x=0\) for \(x<1\) and \(x=2\) for \(x>1\). For the interval \(x<1, (0-1)(0^2+9)>0\) holds true. However, for \(x>1, (2-1)(2^2+9)>0\) is also true which means both intervals satisfy the inequality.
04

Draw the Number Line and Write the Interval Notation

Place the critical point \(x=1\) on your number line. Then, fill in the portions of the line that satisfy the inequality: \((-\infty, 1)\) and \((1,\infty)\). Express the solution set using interval notation which would be \((-\infty, 1) \cup (1,\infty)\). Interval notation uses a round bracket to exclude the endpoint and a square bracket to include it. In this case, the endpoint 1 is excluded from the solution set to the inequality flies open at that point on the graph.

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