What are the critical values of the function y=8secx+4tanx, if 0

To compute the critical values of a function, we'll have
to determine the 1st derivative of the function.

The
solutions of the 1st derivative represent the critical values of the
function.

First, we'll recall the
identities:

sec x = 1/cos
x

tan x = sin x/cos x

We'll
re-write the function:

y = 8/cos x + 4sin x/cos
x

y = (8+4sin x)/cos x

We'll
determine the derivative:

dy/dx = [4(cos x)^2+ 8sin x +
4(sin x)^2]/(cos x)^2

We'll use the Pythagorean identity
(cos x)^2 + (sin x)^2=1

dy/dx = (4+8sinx)/(cos
x)^2

To find the critical values, we'll cancel the 1st
derivative

dy/dx =
0

(4+8sinx)/(cos x)^2 =
0

4+8sin x=0

1+2sin
x=0

2sin x=-1

sin
x=-1/2

The sine function is negative in the 3rd and 4th
quadrants:

x=`pi` + `pi`/6 (3rd
quadrant)

x = 7`pi` /6

x =
2`pi` - `pi` /6

x = 11`pi`
/6

The critical values of the given function,
over the interval (0 ; 2`pi` ) are {7`pi` /6 ; 11`pi`/6
}.