what are critical points for the function g(x)=36x-9tanx
Critical points are zeros, max,min,inflection.
zeros: g(x) = 0: lots of values, such as 0, 1.3, 4.7,... (eyeballing the graph)
max/min: g' = 0: 36 - 9sec^2 = 0
sec^2(x) = 4
sec(x) = 2
x = pi/3, 5pi/3, ...
inflection: g'' = 0: -18secx tanx = 0
where tanx = 0, or
odd multiples of pi/2
Hmmm. Also sec = -2
To find the critical points of a function, we need to find the points where the derivative is either zero or undefined. Let's find the derivative of g(x) first:
g(x) = 36x - 9tan(x)
To compute the derivative, we can use the product rule and chain rule:
g'(x) = 36 - (9sec^2(x))(1)
Now, let's set the derivative g'(x) equal to zero and solve for x:
36 - (9sec^2(x))(1) = 0
36 - 9sec^2(x) = 0
Divide both sides of the equation by 9:
4 - sec^2(x) = 0
Rearrange the equation:
sec^2(x) = 4
Taking the square root of both sides gives:
sec(x) = ±2
Since sec(x) is the reciprocal of cos(x), we have:
cos(x) = ±1/2
To find x, we need to take the inverse cosine (arccos) of both sides:
x = arccos(±1/2)
There are two solutions for arccos(±1/2) within 0 to 2π, which are:
x₁ = π/3
x₂ = 5π/3
So, the critical points of the function g(x) = 36x - 9tan(x) are x = π/3 and x = 5π/3.
To find the critical points of the function g(x) = 36x - 9tan(x), we need to first find its derivative and then solve for values of x where the derivative is either zero or does not exist.
1. Find the derivative of g(x) with respect to x:
g'(x) = 36 - 9(sec^2(x))
2. Set the derivative equal to zero and solve for x:
36 - 9(sec^2(x)) = 0
3. Rearrange the equation:
sec^2(x) = 4
4. Take the square root of both sides:
sec(x) = ±2
5. Since sec(x) is the reciprocal of cos(x), this means that the cosine function should be the reciprocal of 2:
cos(x) = ±1/2
6. Solve for x by taking the inverse cosine (also known as the arccosine function) of both sides:
x = arccos(±1/2)
7. The arccosine of ±1/2 is equal to π/3 and 5π/3, respectively:
x = π/3, 5π/3
8. To check if these values are critical points, substitute them back into the original function g(x) and observe if the derivative changes sign:
When x = π/3:
g'(π/3) = 36 - 9(sec^2(π/3)) = 36 - 9(4) = 0
Since the derivative is zero, x = π/3 is a critical point.
When x = 5π/3:
g'(5π/3) = 36 - 9(sec^2(5π/3)) = 36 - 9(4) = 0
Again, the derivative is zero, so x = 5π/3 is also a critical point.
Therefore, the critical points for the function g(x) = 36x - 9tan(x) are x = π/3 and x = 5π/3.