Question

find the critical numbers of,

f(x)=8secx+4tanx, 0 15 years ago

Answer 1

thanks guys im really freaking out abouit this..lol

Answer 2

sec x = 1/cos x
tan x = sin x/cos x
f(x) = 8/cos x + 4sin x/cos x
f(x) = (8+4sin x)/cos x
We'll determine the derivative:
f'(x) = [4(cos x)^2+ 8sin x + 4(sin x)^2]/(cos x)^2
But (cos x)^2 + (sin x)^2=1
f'(x) = (4+8sinx)/(cos x)^2
To find critical numbers f'(x) = 0
(4+8sinx)/(cos x)^2 = 0
4+8sin x=0
1+2sin x=0
2sin x=-1
sin x=-1/2
x=7pi/6 and x=11pi/6