Finding critical numbers
g(theta) = 16(theta) - 4tan(theta)
I can only get the derivative of g to be g'(theta) = 16 - 4sec^2(theta).
Am I supposed to move on to get sec^2(theta) = 4, then sec(theta) = 2?
g(Ø) = 16Ø - 4tanØ
g ' (Ø) = 16 - 4sec^2 Ø
= 0
4sec^2 Ø = 16
sec^2 Ø = 4
secØ = ±2
cosØ = ± 1/2
we know cos 60° or cos π/3 = 1/2 , and because of the ± , Ø could be in any of the four quadrants
Ø = 60° or 120° or 240° or 300°
or
Ø = π/3, 2π/3 , 4π/3 , 5π/3
To find critical numbers, you need to remember that critical numbers occur when the derivative of a function is either equal to zero or undefined. In this case, since you have already calculated the derivative to be g'(theta) = 16 - 4sec^2(theta), you need to set this derivative equal to zero and solve for theta.
Starting from the equation:
g'(theta) = 16 - 4sec^2(theta) = 0
You can see that the term -4sec^2(theta) equals zero, so you can solve for theta by isolating the sec^2(theta) term:
-4sec^2(theta) = -16
Dividing through by -4:
sec^2(theta) = 4
Now, you're on the right track by considering sec^2(theta) = 4. However, instead of solving for sec(theta), you need to take the square root of both sides to find the possible values of sec(theta):
sqrt(sec^2(theta)) = sqrt(4)
sec(theta) = ±2
Now that you have sec(theta) = ±2, these are the possible values for sec(theta) that would make the derivative equal to zero. By recalling the definition of sec(theta) as the reciprocal of cos(theta), you can conclude that cos(theta) = ±1/2.
To find the values of theta that correspond to these cos(theta) values, you can use the inverse cosine function (cos^(-1)):
cos^(-1)(1/2) ≈ π/3
cos^(-1)(-1/2) ≈ 2π/3
So, the critical numbers occur at theta = π/3 and theta = 2π/3. These are the values of theta where the function g(theta) may have local extrema or inflection points.