M1410207 - (Section 2.7 Nonlinear Inequalities 2.77 SECTION...

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(Section 2.7: Nonlinear Inequalities) 2.77 SECTION 2.7: NONLINEAR INEQUALITIES We solved linear inequalities to find domains, and we discussed intervals in Section 1.4: Notes 1.24 to 1.30 . In this section, we will solve nonlinear inequalities to find domains. Example 1 Let f x ( ) = x 2 9 . We get real outputs x 2 9 0 . There are different ways to solve this inequality; its solution set is the domain of f . Method 1: Sign Chart Method; we are solving x 2 9 0 The key idea here is that we’d rather perform a sign analysis on products of factors as opposed to sums of terms. (For example, the product of a positive real number and a negative real number is guaranteed to be negative; however, there is no such guarantee regarding their sum.) Factoring can be a key tool. x 2 9 0 x + 3 ( ) x 3 ( ) 0 We need to determine where each of the factors on the left side is negative, 0, and positive in value. We ultimately want to know where their product is 0 or positive. The domain of f is: − ∞ , 3 ( 3, ) .
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(Section 2.7: Nonlinear Inequalities) 2.78 Method 2: Parabola Method (for Quadratic Inequalities); we are solving x 2 9 0 The real zeros of x 2 9 are the x -intercepts of the corresponding parabola: x 2 9 = 0 x 2 = 9 x = ± 3 Warning: Here, the ± symbol means “take both the + 3 and the 3 .” See Warning 1 in Section 1.5: Notes 1.46 . The leading coefficient of x 2 9 is positive, so the parabola opens up. This is enough information for us to sketch the parabola to our satisfaction. Given an input x , the y -coordinate of the corresponding point gives the output (or function value). Because of the “ 0 ” in our inequality, we need the values of x , if any, that correspond to the parts of the parabola that lie above or on the x -axis.
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