Answer the following questions for the function f(x)=sin^2(x/5)

defined on the interval .(-15.507923,3.82699075)

Rememer that you can enter "pi" for as part of your answer.

a.what is f(x) concave down on the region

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 what to what and what to what

please help me

First, let's get the derivative of f(x), using the chain rule and the identity for sin(2x)

f' = 2 sin x/5 * cos x/5 * 1/5
= 1/5 sin(2x/5)

OK, a nice simple sine function.

To find where f is concave downward, we need to find where f'' < 0

f'' = 1/5 cos(2x/5) * 2/5
= 2/25 cos(2x/5)

Now, cos(u) < 0 when pi/2 < u < 3pi/2
That is, when
pi/2 < 2x/5 < 3pi/2
or
5pi/4 < x < 15pi/4
or
3.927 < x < 11.781
Hmmm. That's not in our interval. So, subtracting multiples of 2pi*5/2 = 5pi = 15.708 we get
-11.781 < x < -3.297

All the local max/min are also global max/min, since sin^2 oscillates between 0 and 1.

minima are where sin(x/5) = 0, or where x/5 = 0,pi,-pi,...
x = 0, 5pi, -5pi

maxima are where x/5 = pi/2, -pi/2, ...
x = 5pi/2, -5pi/2, ...

f is increasing where f' > 0. That is, where sin(2x/5) > 0
sin(u) > 0 when 0 < u < pi
0 < 2x/5 < pi
0 < x < 5pi/2
-5pi < x < -5pi/2

Plot the graph of f(x) and you can double-check my values...

To determine the answers to the questions, we need to analyze the properties of the function f(x) = sin^2(x/5) on the given interval (-15.507923,3.82699075).

a. Concavity: To find where the function is concave down, we need to find where the second derivative is negative. Let's calculate the second derivative of f(x):

f'(x) = (2/5) * sin(x/5) * cos(x/5)

Using the chain rule, we can calculate the second derivative as follows:

f''(x) = (2/5) * (cos(x/5) * cos(x/5)) - (2/5) * (sin(x/5) * sin(x/5))
= (2/5) * (cos^2(x/5) - sin^2(x/5))

To determine where f(x) is concave down, we need to find where f''(x) < 0. Let's solve the inequality:

(2/5) * (cos^2(x/5) - sin^2(x/5)) < 0

Since the values of cos^2(x/5) and sin^2(x/5) are always between 0 and 1, the overall inequality holds true when cos^2(x/5) < sin^2(x/5). This occurs when cos(x/5) < sin(x/5) or when x/5 is in the range (3pi/4 + 2pi*n, 7pi/4 + 2pi*n), where n is an integer.

Therefore, f(x) is concave down on the intervals (3pi/4 + 2pi*n, 7pi/4 + 2pi*n) within the given range.

b. Global Minimum: To find the global minimum, we need to locate the lowest point on the graph of f(x) within the given interval. Since f(x) = sin^2(x/5), its minimum value is 0, which occurs when sin(x/5) = 0. This happens when x/5 is a multiple of pi, or x = 5n*pi, where n is an integer.

The global minimum for this function occurs at x = 0 in the given interval.

c. Local Maximum: To find the local maximum that is not a global maximum, we need to identify the highest point on the graph of f(x) within the given interval. However, since f(x) = sin^2(x/5) is a bounded function, it does not have a local maximum that is not also a global maximum. Therefore, there is no local maximum for this function that is not a global maximum within the given interval.

d. Increase intervals: To find the intervals where f(x) is increasing, we need to determine where the first derivative is positive.

f'(x) = (2/5) * sin(x/5) * cos(x/5)

Since the sine function oscillates between -1 and 1 for all values of x, we can conclude that sin(x/5) is positive when x/5 is in the range (2pi*n, pi/2 + 2pi*n) or (3pi/2 + 2pi*n, 4pi + 2pi*n), where n is an integer.

Therefore, f(x) is increasing on the intervals (2pi*n, pi/2 + 2pi*n) and (3pi/2 + 2pi*n, 4pi + 2pi*n) within the given range.

Certainly!

To answer these questions, we need to understand the concavity, global minimum, local maximum, and increasing intervals of the function f(x) = sin^2(x/5) on the given interval (-15.507923, 3.82699075).

a. To determine where f(x) is concave down, we need to find the second derivative of the function and analyze its sign.

First, let's find the first derivative of f(x):
f'(x) = 2sin(x/5) * (1/5) * cos(x/5) = (2/5)sin(x/5)cos(x/5)

Now, let's find the second derivative of f(x):
f''(x) = d/dx((2/5)sin(x/5)cos(x/5)) = (2/5) * ((1/5)cos^2(x/5) - (1/5)sin^2(x/5))

To determine when f(x) is concave down, we set f''(x) less than zero and solve for x:
(2/5) * ((1/5)cos^2(x/5) - (1/5)sin^2(x/5)) < 0

Using trigonometric identities, we can simplify this inequality:
cos^2(x/5) - sin^2(x/5) < 0
(cos(x/5) + sin(x/5))(cos(x/5) - sin(x/5)) < 0

The intervals where cos(x/5) + sin(x/5) < 0 and cos(x/5) - sin(x/5) > 0 give us the regions where f(x) is concave down.

b. To find the global minimum of f(x), we need to locate the lowest point on the graph of the function within the given interval.
To do this, we can examine the critical points and endpoints of the interval.

To find critical points, we set the first derivative f'(x) equal to zero:
(2/5)sin(x/5)cos(x/5) = 0

This equation is satisfied when sin(x/5) = 0 or cos(x/5) = 0.
For sin(x/5) = 0, x/5 = kπ where k is an integer. Therefore, x = 5kπ.
For cos(x/5) = 0, x/5 = (k + 1/2)π where k is an integer. Therefore, x = (5k + 2/5)π.

Now, we need to check each critical value and the endpoints of the interval to determine the global minimum.

c. To find a local maximum of f(x) that is not a global maximum, we need to locate the highest point on the graph of the function within the given interval.
Again, we can examine the critical points and endpoints of the interval to find this local maximum.

d. To determine the intervals where f(x) is increasing, we need to find where the first derivative f'(x) is positive.
To do this, we set f'(x) greater than zero and solve for x:
(2/5)sin(x/5)cos(x/5) > 0

Using the same approach as in part a, we can find the intervals where f(x) is increasing.

These are the general steps to find the concavity, global minimum, local maximum, and increasing intervals for the given function on the provided interval. I recommend using a graphing calculator or graphing software to visualize the function and check your answers.