What are the critical values of the function y=8secx+4tanx, if 0<x<2pi?

Expert Answers
justaguide eNotes educator| Certified Educator

The critical values of a function are those at which the slope of the curve representing the function changes sign. These are located at the points where the first derivative is equal to 0.

y = 8*sec x + 4*tan x

y' = 8*sec x* tan x + 4*(sec x)^2

8*sec x* tan x + 4*(sec x)^2 = 0

=> 4*sec x[2*tan x + sec x] = 0

sec x = 0 has no solution for x.

2*tan x + sec x = 0

=> 2*sin x/cos x + 1/cos x = 0

=> sin x = -1/2

x = 330 degrees and x = 210 degrees

The critical points are at x = 210 degrees and x = 330 degrees

giorgiana1976 | Student

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 }.