Calculus math problem?

Question

i'm really stuggling with this problem how do i solve it?

f(x)= 3 sec(2x) + 1

a.determine the period.

b.determine the domain.

c.determine the equations of the vertical asymptotes.

d.determine the vertical shift.

e.write the coordinates of two local max. and min points.

Answer 1

(a)

All the basic trig functions have period 2pi.

If you put 2x = u, sec(2x) becomes sec(u), which has period 2pi for u, and therefore period pi for x.

(b) and (c)

sec(u) = 1 / cos(u).

cos(u) can have all values in [- 1, 1].

Some of those values are 0, namely where u = pi / 2, 3pi / 2, 5pi / 2 and so on.

These points can therefore not be in the domain of sec(u).

sec(2x) has vertical assmptotes at these points, and therefore sec(2x) has vertical asymptotes at pi / 4, 3pi / 4, 5pi / 4 ...

Its domain is:

(- infty, infty) excluding { (2n + 1) pi / 4 } where n is any integer.

(d)

The vertical shift is 1, as every point on the graph y = f(x) is 1 higher than the corresponding point on the graph of y = 3 sec(2x).

(e)

The maxima of sec(2x) correspond to the points where cos(2x) = - 1, and the minima to the points where cos(2x) = 1.

That gives maxima 2x = (2n + 1) pi, or x = (2n + 1)pi / 2; minima 2x = 2n pi, or x = n pi.

Two local maxima:

(pi / 2, - 2), (3pi / 2, - 2).

Two local minima:

(0, 4), (pi, 4).

Source(s):

You can plot the graph here:

http://www.walterzorn.com/grapher/graphe...