Stitz-Zeager_College_Algebra_e-book

01 which is geometrically impossible next we have an

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

Ask a homework question - tutors are online