Critical numbers occur where the first derivative is 0 or undefined and is in the domain of the original function...so let's take the derivative of the function using power rule: d/dx (ax^b) = (a*b)x^(b-1) and that derivative of tan x is sec^2 x...so..
g(θ) = 36θ - 9 tan θ
g'(θ) = 36 * 1θ^(1-1) - 9 * sec^2(θ)
g'(θ) = 36 - 9 sec^2 (θ)
Now notice that sec^2 θ = 1/(cos^2 θ) so the first derivative is undefined at cos^2 θ = 0 but those values of x are NOT in the domain of the original function so they are not critical numbers..
Now look where the first derivative is 0 is where there is critical numbers...
0 = 36 - 9 sec^2 θ
9 sec^2 θ = 36 - 9 sec^2 θ + 9 sec^2 θ
9 sec^2 θ = 36
(9 sec^2 θ)/9 = 36/9
sec^2 θ = 4 <-- rewrite in terms of cosine...
1/(cos^2 θ) = 4 <-- multiply both sides of the equation by cos^2 θ to get rid of fraction...
1 = 4 cos^2 θ <-- divide both sides by 4 to isolate cos^2 θ...
1/4 = cos^2 θ <-- take square root of both sides of the equation...
±1/2 = cos (θ)
So cos θ = 1/2 (cosine is positive in quadrant I and IV, find x-coordinates where it is 1/2 so that it is and then add the period of the cosine onto that angle)...
It is where θ = π/3 + 2nπ and...(2π - π/3) + 2nπ = 5π/3 + 2nπ....
Where cos θ = -1/2 is (cosine is negative in quadrant II and III so find where x-coordinate is -1/2 and add on the period of cosine) is at θ = (π - π/3) + 2nπ = 2π/3 + 2nπ and (π + π/3) + 2nπ = 4π/3 + 2nπ where n is an integer...
Answer: θ = π/3 + 2nπ, 2π/3 + 2nπ, 4π/3 + 2nπ and 5π/3 + 2nπ where n is an integer...
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