Use the quadratic equation y = x2 + 6x + 36.
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g(θ) = 20θ − 5 tan θ
To find out critical points we take first derivative and set it =0
g(θ) = 20θ − 5 tan θ
g'(θ) = 20 − 5 sec^2(θ)
Now we set derivative =0
20 − 5 sec^2(θ)=0
Subtract 20 from both sides
− 5 sec^2(θ)=0 -20
Divide both sides by 5
sec^2(θ)= 4
Take square root on both sides
sec(θ)= -2 and sec(θ)= +2
sec can be written as 1/cos
so sec(θ)= -2 can be written as cos(θ)= -1/2
Using unit circle the value of θ is
sec(θ)= 2 can be written as cos(θ)=1/2
Using unit circle the value of θ is
For general solution we add 2npi
So critical points are