# 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]?

Apr 22, 2016

$\left(0 , \frac{2 \pi}{3}\right]$, open on the left and closed on the right.

#### Explanation:

f(x) is periodic with period $2 \pi$.

The half of a full wave, in $\left(0 , \pi\right)$, is up, up to 8 units at $x = \frac{\pi}{2}$, while the shorter half, in $\left(\pi , 2 \pi\right)$ goes down up to -4 units at $x = \frac{3 \pi}{2}$.

The tangent crosses the wave at $x = 0 , \pi , 2 \pi$, 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 $\left[\arcsin \left(\frac{- 5 + \sqrt{57}}{8}\right) , \frac{2 \pi}{3}\right]$.

Which is about $\left[0.3244 , \frac{2 \pi}{3}\right]$.

#### Explanation:

$f ' \left(x\right) = 5 \cos x + 2 \sin x \cos x = 5 \cos x + \sin \left(2 x\right)$

$f ' ' \left(x\right) = - 5 \sin x + 2 \cos \left(2 x\right) = - 5 \sin x + 1 \left(1 - 2 {\sin}^{2} x\right)$

$f ' ' \left(x\right) = - 4 {\sin}^{2} x - 5 \sin x + 2$

Solving $f ' ' \left(x\right) = 0$ we find the partition numbers using

$\sin x = \frac{- 5 + \sqrt{57}}{8}$

The only value of $x$ in the interval with this sine is $\arcsin \left(\frac{- 5 + \sqrt{57}}{8}\right) \approx 0.3244$

For $- \frac{\pi}{3} < x < 0.3244$, we have $f ' ' \left(x\right) > 0$, so $f$ is concave up.
(Use $x = 0$ as a test number.)

For $0.3244 < x < \frac{2 \pi}{3}$, we have $f ' ' \left(x\right) < 0$, so $f$ is concave down.
(Use $x = \frac{\pi}{2}$ as a test number.)

So the largest interval is the closed interval $\left[\arcsin \left(\frac{- 5 + \sqrt{57}}{8}\right) , \frac{2 \pi}{3}\right]$.

Which is about $\left[0.3244 , \frac{2 \pi}{3}\right]$.