Question 1014764
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Analyze the function f(x) = sec 2x. Include:
- Domain and range
- Period and Amplitude
- Two Vertical Asymptotes
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f(x) = sec(2x) = {{{1/cos(2x)}}}.


The domain is all number line (all real numbers) except those where cos(2x) = 0. 


These exceptional values are 2x = {{{pi/2 + k*pi}}}, k = 0, =/-1, +/-2, . . . 

In other words, the exceptional values for x are {{{pi/4 + k*(pi/2)}}}, k = 0, =/-1, +/-2, . . . 


The range is the union of two infinite intervals: y <= -1  and  y >=1.


The period is the same as for the function cos(2x), i.e. {{{pi}}}.


There is no amplitude. The conception/notion of an amplitude is not applicable to the function y = sec(2x).


Vertical asymptotes are there where cos(2x) = 0, i.e. at the points {{{pi/4 + k*(pi/2)}}}, k = 0, =/-1, +/-2, . . . - the same exceptional points mentioned above.


Plot of the function y = sec(2x) is shown below.

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  <TD> 

{{{graph( 330, 330, -8.5, 8.5, -10.5, 10.5,
          1/cos(2x)
)}}}


<B>Figure</B>. Plot y = sec(2x)

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