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Stitz-Zeager_College_Algebra_e-book

# Theorem 116 herons formula suppose a b and c denote

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Unformatted text preview: = tan(arctan(117)) − 3 = 114. We choose our π π last test value to be x = 74 and ﬁnd f 74 = −4. Since we want f (x) ≥ 0, we see that our π answer is arctan(3), π ∪ arctan(3) + π, 32 . Using the graphs of y = tan(x) and y = 3, we 2 can verify when the graph of the former is above (or meets) the graph of the latter. 10 See page 247 for a discussion of the non-standard character known as the interrobang. We could have chosen any value arctan(t) where t > 3. 12 . . . by adding π through the inequality . . . 11 10.7 Trigonometric Equations and Inequalities (−) 0 (+) 0 (−) 0 (+) π 2 arctan(3) (arctan(3) + π ) 739 (−) 3π 2 2π y = tan(x) and y = 3 We close this section with an example that puts solving equations and inequalities to good use – ﬁnding domains of functions. Example 10.7.4. Express the domain of the following functions using extended interval notation.13 1. f (x) = csc 2x + π 3 2. f (x) = sin(x) 2 cos(x) − 1 3. f (x) = 1 − cot(x) Solution. 1. To ﬁnd the domain of f (x) = csc 2x + π 3 , we rewrite f in terms of sine as f (x) = 1 . sin(2x+ π ) 3 Since the sine function is deﬁned everywhere, our only concern comes from zeros in the denominator. Solving sin 2x + π = 0, we get...
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