Question

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]`?

Answer 1

`(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..

Answer 2

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]`.