Stitz-Zeager_College_Algebra_e-book

2 6 3 checking our answers we have that for any

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Unformatted text preview: , and sure enough, we run into trouble 2 π at x = π and x = 32 since cos(x) = 0 at these values. Using the notation introduced in Section 2 4.2, we have that as x → π − , cos(x) → 0+ , so sec(x) → ∞.11 Similarly, we find that as x → π + , 2 2 π− π+ sec(x) → −∞, as x → 32 , sec(x) → −∞, and as x → 32 , sec(x) → ∞. This means we have a π pair of vertical asymptotes to the graph of y = sec(x), x = π and x = 32 . Since cos(x) is periodic 2 with period 2π , it follows that sec(x) is also.12 Below we graph a fundamental cycle of y = sec(x) along with a more complete graph obtained by the usual ‘copying and pasting.’13 y x 0 cos(x) 1 π 4 π 2 3π 4 2 2 √ 0 undefined √ √ − 22 −2 π −1 5π 4 3π 2 7π 4 2 2 2π sec(x) 1 √ 2 √ − −1 √ −2 0 undefined √ √ 2 2 2 1 1 3 (x, sec(x)) (0, 1) √ π 4, 2 √ 3π 4 ,− 2 1 2 π 4 (π, −1) √ 2 π 2 3π 4 π 5π 4 3π 2 7π 4 x 2π −1 5π 4 ,− −2 7π 4, √ 2 −3 (2π, 1) The ‘fundamental cycle’...
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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.

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