Calc. Concave Down?

Question

I need help finding where the fuction f(x)=sin^2(x/5) concaves down. What regions does f(x) concave down ___ to ____

on the intervals [-15.207963, 2.92699075]

Answer 1

f(x) is concave down when f''(x) < 0.

f'(x) = 2 sin(x/5) cos(x/5) (1/5) = (1/5) sin(2x/5), using the formula sin(2t) = 2 sin(t) cos(t).

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

cos(t) < 0 when pi/2 + 2k*pi < t < 3pi/2 + 2k*pi, for any integer k. Therefore f is concave down when

(5/2)(pi/2 + 2k*pi) < x < (5/2)(3pi/2 + 2k*pi)
or
5pi/4 + 5k*pi < x < 15pi/4 + 5k*pi

Set k = -1 for values within the specified interval.

Answer 2

"Concave down" is where acceleration is negative. You take the double differential, or d(dF/dx)/dx, and the "concave down" places are where it's negative. So, the double differential of sin^2(x/5) is 2/25 Cos(2x/5), and so you can see that the negative values begin where 2x/5 = pi/2, etc.

Answer 3

f'(x) = 2 sin(x/5) cos(x/5) (1/5) = (1/5) sin(2x/5),
f''(x) = (1/5) cos(2x/5) (2/5) = (2/25) cos(2x/5).
f''(x)=0 if 2x/5 =... -3pi/2, -pi/.2, pi/2, 3pi/2 ....
so you need to check between those points
in particular
for 2x/5 between pi/2 and 3pi/2 it will be negative
x is between 5pi and 15pi etc etc etc
s