Answer the following questions for the function
f(x) = sin^2(x/3)
defined on the interval [ -9.424778, 2.356194].
Rememer that you can enter pi for \pi as part of your answer.
a.) f(x) is concave down on the interval .
I'm really stuck on this one question, can't seem to find the interval.
f = sin^2(x/3)
f' = 2/3 sin(x/3)cos(x/3) = 1/3 sin(2x/3)
f'' = 2/9 cos(2x/3)
f is concave down when f'' < 0
so, where is cos(2x/3) < 0?
When pi/2 < 2x/3 < 3pi/2
or 3pi/4 < x < 9pi/4
Your interval appears to be about (-3pi/2,pi/2), so you may have to adjust your answer a bit to fit the interval.
To determine where the function f(x) = sin^2(x/3) is concave down, we need to find the interval where the second derivative of the function is negative.
First, let's find the derivative of f(x). The derivative of sin^2(x/3) can be found by applying the chain rule.
f'(x) = 2sin(x/3) * (1/3) * cos(x/3)
= (2/3)sin(x/3)cos(x/3)
The second derivative can be found by differentiating f'(x) with respect to x:
f''(x) = d/dx [(2/3)sin(x/3)cos(x/3)]
= (2/3)[cos^2(x/3) - sin^2(x/3)]
To find the interval where f(x) is concave down, we need to solve the inequality f''(x) < 0.
(2/3)[cos^2(x/3) - sin^2(x/3)] < 0
Next, we simplify the inequality by factoring out (2/3):
cos^2(x/3) - sin^2(x/3) < 0
Now, we can use the trigonometric identity cos^2(x) - sin^2(x) = cos(2x) to rewrite the inequality:
cos(2(x/3)) < 0
The inequality cos(2(x/3)) < 0 is satisfied when the cosine function is negative. This occurs in the intervals where x/3 lies between π/2 and 3π/2 (i.e., the second and third quadrants).
Multiplying by 3, we get:
x < 3π/2 and x > π/2
Since we are given the interval [-9.424778, 2.356194], we need to find the intersection of this interval with the solution of the inequality.
The interval [π/2, 3π/2] is [1.570796, 4.712389].
Hence, the interval where f(x) is concave down is the intersection of [-9.424778, 2.356194] and [1.570796, 4.712389], which is [1.570796, 2.356194].
Therefore, f(x) is concave down on the interval [1.570796, 2.356194].