# What are the points of inflection of f(x)= 2sin (x) - cos^2 (x) on x in [0, 2pi] ?

Jul 28, 2018

They are $\left(\frac{\pi}{6} , \frac{1}{4}\right)$ and $\left(\frac{5 \pi}{6} , \frac{1}{4}\right)$

#### Explanation:

$f \left(x\right) = 2 \sin x - {\cos}^{2} x$
$\Rightarrow f ' \left(x\right) = 2 \cos x - 2 \cdot \cos x \cdot \sin x$ (by the chain rule).
$= 2 \cos x - \sin \left(2 x\right)$ (by the double-angle formula for $\sin$).
$\Rightarrow f ' ' \left(x\right) = - 2 \sin x - 2 \cos \left(2 x\right)$
$= - 2 \left(\sin x - \cos \left(2 x\right)\right)$

One of the conditions for a point of inflexion at $x = a$ is that $f ' ' \left(a\right) = 0$ ... however, $f ' ' \left(x\right)$ must also change sign.

First we solve the equation $f ' ' \left(x\right) = 0$
$= - 2 \left(\sin x - \cos \left(2 x\right)\right) = 0$
$\Rightarrow \sin x - \cos \left(2 x\right) = 0$
Expanding $\cos \left(2 x\right)$ with one of the double-angle formulas for $\cos$:
$\sin x - \left(1 - 2 {\sin}^{2} x\right) = 0$
$\therefore 2 {\sin}^{2} x + \sin x - 1 = 0$

Now we let $u = \sin x$
$2 {u}^{2} + u - 1 = 0$
$\Rightarrow u = - 1$ or $u = \frac{1}{2}$ (by factorising the quadratic).
$\Rightarrow x \in \left\{\frac{\pi}{6} , \frac{5 \pi}{6} , \frac{3 \pi}{2}\right\}$ (taking the inverse sine of the previous $u$-values in the given domain).

Now, to ensure that $f ' ' \left(x\right)$ changes sign across these $x$-values, we can just evaluate the following:
$f ' ' \left(0\right) = 2$
$f ' ' \left(\frac{\pi}{2}\right) = - 4$
$f ' ' \left(\pi\right) = 2$
$f ' ' \left(2 \pi\right) = 2$

This shows that $f ' ' \left(x\right)$ changes sign across $x = \frac{\pi}{6}$ and $x = \frac{5 \pi}{6}$ but not $x = \frac{3 \pi}{2}$, so the there is no point of inflexion at $x = \frac{3 \pi}{2}$

So, the co-ordinates of the points of inflexion for $f \left(x\right)$ are:
$\left(\frac{\pi}{6} , \frac{1}{4}\right)$ and $\left(\frac{5 \pi}{6} , \frac{1}{4}\right)$

Hope this helps!