Find the critical numbers of the function on the interval 0 ≤ θ < 2π. g(θ) = 4 θ - tan(θ)?
I found the smallest number to be 1.047, and the second to largest number to be 4.189. I need to know the critical number for the tow asymptotes on the equations graph.
Answers & Comments
Verified answer
First, I learned you had to derive that first
so if you derive the function, it is
g'(x) = 4-sec^2(x) right?
and we had to make that equal to 0 so
0 = 4-sec^2(x)
now solve for x
sec^2(x) = 4
square root both side and you get
sec(x) = +- 2
and we know sec(x) = 1/cos(x)
so 1/cos(x) = +- 2
From here, make it in fraction form
1/cos(x) = 2/1 and cross multiply
1 = 2cos(x)
now divide both sides by 2 and you get
1/2 = cos(x)
for what angle (x) will cos(x) equal 1/2?, angles are 60 and 300
do the same thing I did above but this time with -2
1 = -2cos(x)
-1/2 = cos(x)
Angles(x) for cos(x) to equal -1/2 are 120 and 240
now we have angles 60, 120, 240 and 300 for the x value.
We know from pre-calc that Pi = 3.14 or angle-wise, 180 degrees
Now we know the pi and the radian form and that 60 = pi / 3
if you do this on your calculator, pi/3 = 3.14/3 which should be 1.047 which is what you got for your lowest value
120 = 2pi/3 = 2.094
240 = 4pi/3 = 4.189 (what you got)
300 = 5pi/3 = 5.236
So answers are 1.047, 2.904, 4.189, 5.236 in this order.
I think what you didn't do is when you square rooted sec^2(x) = 4
you didn't think of -2 and did only sec(x) = 2
that's how you got angle 60, and 240, but not 120 and 300 which are -1/2 value for cos(x)
Hope this helped.