f(x) = sin^2(x/2)
defined on the interval [ -5.683185, 1.270796].
Remember that you can enter pi for \pi 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 .
f = sin^2(x/2)
f' = 2sin(x/2)cos(x/2)(1/2) = sin(x)/2
f'' = cos(x)/2
concave down where f'' < 0: [-3pi/2,-pi/2]
global min where f'=0, f''>0: 0
local max where f'=0, f''<0: -pi
This is also the global max
f is increasing where f'>0: [-5.683185,-pi)U(0,1.270796]
Technically,
concave down where f'' < 0: (-3pi/2,-pi/2)
thank you, that really help me
To determine the concavity of a function, we need to find the second derivative and analyze its sign.
a) To find the concavity of f(x), we need to calculate the second derivative. The first step is to find the first derivative:
f(x) = sin^2(x/2)
f'(x) = 2*sin(x/2)*cos(x/2) (using the chain rule)
Next, we find the second derivative:
f''(x) = [2*cos(x/2)*cos(x/2)] - [2*sin(x/2)*sin(x/2)]
f''(x) = 2*cos^2(x/2) - 2*sin^2(x/2)
f''(x) = 2(cos^2(x/2) - sin^2(x/2))
Now we need to analyze the sign of f''(x) on the given interval [-5.683185, 1.270796]. To do this, we can choose sample points within the interval and evaluate the second derivative at those points. We'll use x = -4 and x = 1 as sample points.
When we substitute x = -4 into f''(x), we get:
f''(-4) = 2(cos^2(-4/2) - sin^2(-4/2))
f''(-4) = 2(cos^2(-2) - sin^2(-2))
Similarly, when we substitute x = 1 into f''(x), we get:
f''(1) = 2(cos^2(1/2) - sin^2(1/2))
By evaluating f''(-4) and f''(1), we can determine the sign of f''(x) on the interval [-5.683185, 1.270796]. If f''(x) is positive, the function is concave up. If f''(x) is negative, the function is concave down.
b) To find the global minimum of the function, we need to locate the critical points. Critical points occur where the first derivative is equal to zero or is undefined. In our case, the critical points are the values of x for which f'(x) = 0 or f'(x) is undefined.
Setting f'(x) = 0, we get:
2*sin(x/2)*cos(x/2) = 0
Simplifying this equation, we find:
sin(x/2)*cos(x/2) = 0
This equation is true when either sin(x/2) = 0 or cos(x/2) = 0.
For sin(x/2) = 0, we have x/2 = n*pi, where n is an integer. Solving for x, we get:
x = 2n*pi
For cos(x/2) = 0, we have x/2 = (2n+1)*pi/2, where n is an integer. Solving for x, we get:
x = (2n+1)*pi
We now need to check if these critical points lie within the given interval [-5.683185, 1.270796]. If they do, we evaluate f(x) at each critical point and compare the values to find the global minimum.
c) To find a local maximum that is not a global maximum, we need to look for critical points within the given interval [-5.683185, 1.270796] where f''(x) changes sign. These points will correspond to local extrema.
Similar to finding the critical points in part b, we set f'(x) = 0 and solve for x. However, this time we also consider the points where f''(x) is undefined. We then check if these critical points lie within the given interval and determine their corresponding y-values to find the local maximum that is not a global maximum.
d) To determine where the function is increasing, we need to analyze the sign of the first derivative, f'(x), on the given interval [-5.683185, 1.270796]. If f'(x) is positive, the function is increasing. If f'(x) is negative, the function is decreasing.
By examining the sign of f'(x) at specific sample points within the given interval, we can determine where the function is increasing.