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

You need to differentiate the equation to find the answer
Answer:y = - 2x + 7/6
Step-by-step explanation:
The point slope form of the equation is expressed as
y - 2/3 = - 2(x - 1/4)
Comparing with the point slope form of an equation,
y - 1 = m(x - 1)
From the given equation,
Slope, m= - 2
The point through which the line passes through is ( 1/4 , 2/3)
The slope intercept form is expressed as
y = mx + c
Where
m represents slope
c represents intercept
To determine c, we we would substitute m = - 2, x = 1/4 and y = 2/3 into y = mx + c. It becomes
2/3 = - 2 × 1/4 + c
2/3 = - 1/2 + c
c = 2/3 + 1/2 = 7/6
Therefore, the slope intercept form would be
y = - 2x + 7/6
Answer:
These should be right hopefully! Hope it helps!