Answer the following questions for the function
f(x)=sin(x/4)^2
defined on the interval [-12.266371, 2.241593].
Rememer that you can enter pi for as part of your answer.
a.)f(x) is concave down on the interval ____.
b.) A global minimum for this function occurs at ____.
c.) A local maximum for this function which is not a global
maximum occurs at _____.
d.) The function is increasing on the region _____.
I REALLY NEED HELP ANSWERING THIS. I mange to get b and c but i'm stuck on the other two
What did you get for b & c?
For concavity and increasing, you need to calculate f"(x) within the interval.
Concavity changes when f"(x)=0. f"(x)>0 is concave upwards, and vice versa.
Function's and increasing/decreasing changes only when f'(x)=0. When f'(x)>0 it is increasing, and vice versa.
To determine the concavity of the function f(x)=sin(x/4)^2, we need to find the second derivative of the function. The concavity of a function is determined by the sign of the second derivative.
a.) To find where the function is concave down, we need to find the interval where the second derivative is negative. Let's start by finding the first derivative:
f'(x) = (2/4)sin(x/4) * (1/4) = (1/8)sin(x/4)
Now, let's find the second derivative:
f''(x) = (1/8)cos(x/4) * (1/4) = (1/32)cos(x/4)
To determine where the function is concave down, set the second derivative less than zero:
(1/32)cos(x/4) < 0
Since the cosine function is negative in the second and third quadrants, we need to find where x/4 falls in those quadrants. To find the interval, we can set the argument of the cosine function equal to the appropriate angles:
x/4 = 3pi/2 + 2kpi and x/4 = 2pi + 2kpi, where k is an integer.
Multiplying both sides by 4 to solve for x:
x = 6pi + 8kpi and x = 8pi + 8kpi
The interval where the function f(x) is concave down is:
[-12pi/4, 6pi/4] = [-3pi, 3pi]
b.) To find the global minimum of the function, we need to determine the critical points.
First, let's set the derivative f'(x) to zero and solve for x:
(1/8)sin(x/4) = 0
Sin(x/4) is zero when x/4 is a multiple of pi:
x/4 = kpi
Multiplying both sides by 4 to solve for x:
x = 4kpi
Now, we need to check the interval [-12.266371, 2.241593] to see if any of the critical points fall in that range.
Substituting the boundary values into the equation x = 4kpi:
-12.266371 = 4kpi and 2.241593 = 4kpi
Solving for k in each equation:
k = -12.266371 / (4pi) and k = 2.241593 / (4pi)
The value of k should be rounded to the nearest integer since k is an integer in x = 4kpi.
The global minimum occurs at the x-value 4kpi that is within the given interval. The value of k will depend on the boundary value you choose to use (either -12.266371 or 2.241593):
Global Minimum = 4 * rounded k * pi
c.) To find a local maximum that is not a global maximum, we need to look for any local maximum points other than the global minimum we found in part b.
We can do this by finding the critical points where the derivative changes sign or by finding the second derivative.
To find the critical points, set the derivative f'(x) to zero and solve for x:
(1/8)sin(x/4) = 0
Similar to part b, we get the equation x/4 = kpi, and solving for x gives x = 4kpi.
Check the interval [-12.266371, 2.241593] for any local maximum points within that range.
Substituting the boundary values into the equation x = 4kpi:
-12.266371 = 4kpi and 2.241593 = 4kpi
Solving for k in each equation:
k = -12.266371 / (4pi) and k = 2.241593 / (4pi)
The value of k should be rounded to the nearest integer since k is an integer in x = 4kpi.
The local maximum that is not a global maximum occurs at the x-value 4kpi that is within the given interval. The value of k will depend on the boundary value you choose to use (either -12.266371 or 2.241593):
Local Maximum = 4 * rounded k * pi
d.) To determine where the function is increasing, we need to find the intervals where the derivative is positive.
From previous calculations, we found that:
f'(x) = (1/8)sin(x/4)
To find where f'(x) > 0, we need to find where sin(x/4) > 0.
Since sin(x/4) is positive in the first and second quadrants, we need to find where x/4 falls in those quadrants.
To find the interval, we can set the argument of the sine function equal to the appropriate angles:
x/4 = pi/2 + 2kpi and x/4 = pi + 2kpi, where k is an integer.
Multiplying both sides by 4 to solve for x:
x = 2pi + 8kpi and x = 4pi + 8kpi
The function f(x) is increasing on the region:
[2pi + 8kpi, 4pi + 8kpi]