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.
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.
1 answer