# Analyze the function f(x) = sec (2x)?

Analyze the function f(x) = sec (2x).

Give the following:

- Domain and range

- Period and Amplitude

- Two Vertical Asymptotes

Relevance

sec (2x) = 1/cos(2x)

The period of cos (2x) can be found using the formula

Period = 2pi / B

In this case B = 2. So the period of cos(2x) = period of sec(2x) = 2pi/2 = pi or 180 degrees

The amplitude is defined to be the constant in front of the trg function. For sec(2x) this constant is 1.

The domain is where sec(2x) is defined. This will be everywhere that cos(2x) does not equal 0.

cos(x) is zero at 90 +/- 360*n where n is an integer. (you can convert to radians if you want)

It is also zero at 270 +/- 360*n where n is an integer.

cos(2x) is zero twice as often.

So the domain of cos (2x) =

cos(x) is zero at 45 +/- 180*n where n is an integer.

It is also zero at 135 +/- 180*n where n is an integer.

The range of sec(2x) is 1 over the range of cos(2x) so

1 <= x < infinity and

-1 => x > minus infinity

Two vertical asymptotes occur where sec(2x) is not defined so 45 and 135 (of course there are an infinite number of them.)

Hiope this helps.