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Stitz-Zeager_College_Algebra_e-book

# 01 which is geometrically impossible next we have an

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Unformatted text preview: mer on the sign diagram.10 Moving along to zeros, solving f (x) = tan(x) − 3 = 0 requires the arctangent function. We ﬁnd x = arctan(3) + πk for integers k and of these, only x = arctan(3) and x = arctan(3) + π lie in [0, 2π ). Since 3 > 0, we know 0 < arctan(3) < π which allows us to position these zeros correctly on the 2 sign diagram. To choose test values, we begin with x = 0 and ﬁnd f (0) = −3. Finding a convenient test value in the interval arctan(3), π is a bit more challenging. Keep in mind 2 that the arctangent function is increasing and is bounded above by π . This means the number 2 x = arctan(117) is guaranteed to lie between arctan(3) and π .11 We ﬁnd f (arctan(117)) = 2 tan(arctan(117)) − 3 = 114. For our next test value, we take x = π and ﬁnd f (π ) = −3. To ﬁnd our next test value, we note that since arctan(3) < arctan(117) < π , it follows12 2 π that arctan(3) + π < arctan(117) + π < 32 . Evaluating f at x = arctan(117) + π yields f (arctan(117) + π ) = tan(arctan(117) + π ) − 3...
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