9514 1404 393
Answer:
a = 36
Step-by-step explanation:
A stationary point is where the function derivative is zero. Here, we have ...
y' = 2kx -16/x^2 = 0
For x = 2/3, this is ...
4/3k -16/(4/9) = 0
4/3k = 36
k = 27
Then the value of y at that point is ...
y = 27(2/3)^2 + 16/(2/3) = 12 +24 = 36
The stationary point of interest is (2/3, a) = (2/3, 36). a = 36.
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
