Question

Question #56aac

Answer 1

Local minimum is 1 at `x = 2kpi, k = 0, +-1, +-2, +-3,...`
Local maximum is -1 at `x = (2k+1)pi, k = 0, +-1, +-2, +-3,...`
There is no global maximum nor minimm.

Explanation:

Here, `y = sec x = 1/cos x`.

For `x in (0, pi/2), y in (1, oo)`

For `x in (pi/2, pi], y in (-oo, -1)`

For `x in [pi, 3/2pi), y in (-1, -oo)`

For `x in (3/2pi, 2pi), y in (oo, 1)`

The period of sec x being `2pi`, the ranges repeat this way,

for `x in (2kpi, 2(k+2)pi), k = 0, +-1, +-2, +-3, ...`

Note the infinite discontinuities at `x = pi/2 and 3/2pi`

Local minimum is 1 at `x = 2kpi, k = 0, +-1, +-2, +-3,...`

Local maximum is -1 at `x = (2k+1)pi, k = 0, +-1, +-2, +-3,...`

There is no global maximum nor minimm