What is the largest interval on which the function is concave down for #f(x) = 5sin(x) + (sin(x))^2# over the interval #[-pi/3, 2pi/3]#?

2 Answers
Apr 22, 2016

#(0, (2pi)/3]#, open on the left and closed on the right.

Explanation:

f(x) is periodic with period #2pi#.

The half of a full wave, in #(0, pi)#, is up, up to 8 units at #x = pi/2#, while the shorter half, in #(pi, 2pi)# goes down up to -4 units at #x = (3pi)/2#.

The tangent crosses the wave at #x=0, pi, 2pi#, to turn the curve from concavity to convexity, and vice versa. Of course, viewing from the x-axis, it is looking concave, both sides.

I think you would say OK to my answer now..

Apr 22, 2016

If you need exact values, I got #[arcsin((-5+sqrt57)/8),(2pi)/3]#.

Which is about #[0.3244, (2pi)/3]#.

Explanation:

#f'(x) = 5cosx+2sinxcosx = 5cosx+sin(2x)#

#f''(x) = -5sinx+2cos(2x) = -5sinx+1(1-2sin^2x)#

#f''(x) = -4sin^2x-5sinx+2#

Solving #f''(x) = 0# we find the partition numbers using

#sinx = (-5+sqrt57)/8#

The only value of #x# in the interval with this sine is #arcsin((-5+sqrt57)/8) ~~0.3244#

For #-pi/3 < x < 0.3244#, we have #f''(x) > 0#, so #f# is concave up.
(Use #x=0# as a test number.)

For # 0.3244 < x < (2pi)/3#, we have #f''(x) < 0#, so #f# is concave down.
(Use #x=pi/2# as a test number.)

So the largest interval is the closed interval #[arcsin((-5+sqrt57)/8),(2pi)/3]#.

Which is about #[0.3244, (2pi)/3]#.