# How do you find all local maximum and minimum points using the second derivative test given y=sin^3x?

Nov 8, 2016

Points of Inflection occur when $x = n \pi$,
ie $x = - 3 \pi , - 2 \pi , - \pi , 0 , \pi , 2 \pi , 3 \pi , \ldots$

Maximum points occur when $x = n \pi - \frac{\pi}{2}$ and n is odd
ie $x = \frac{- 11 \pi}{2} , \frac{- 7 \pi}{2} , \frac{- 3 \pi}{2} , \frac{\pi}{2} , \frac{5 \pi}{2} , \frac{9 \pi}{2} , \ldots$

Minimum points occur when $x = n \pi - \frac{\pi}{2}$ and n is even
$x = \frac{- 9 \pi}{2} , \frac{- 5 \pi}{2} , \left(- \frac{\pi}{2}\right) , \frac{3 \pi}{2} , \frac{7 \pi}{2} , \ldots$

#### Explanation:

We have $y = {\sin}^{3} x$ .... [1]

We first find the max/min critical points by finding values of $x$ such that $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Differentiating [1] wrt $x$ using the chain rule we get:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {\sin}^{2} x \cos x$ .... [2]

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 3 {\sin}^{2} x \cos x = 0$
$\therefore {\sin}^{2} x \cos x = 0$
$\therefore {\sin}^{2} x = 0$ or $\cos x = 0$

If ${\sin}^{2} x = 0 \implies \sin x = 0 \implies x = n \pi , n \in \mathbb{Z}$
And $\cos x = 0 \implies x = n \pi - \frac{\pi}{2} , n \in \mathbb{Z}$

To determine the nature of these points we need to look at the second derivative

Differentiating [2] wrt [x] and simultaneously applying the chain rule and product rule we have;
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \left(3 {\sin}^{2} x\right) \left(- \sin x\right) + \left(6 \sin x \cos x\right) \left(\cos x\right)$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 \sin x {\cos}^{2} x - 3 {\sin}^{3} x$

We already know that min/max occurs when $x = n \pi$ or$x = n \pi - \frac{\pi}{2}$, so let's find $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$ at these points;

When $x = n \pi \implies \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 \left(0\right) {\cos}^{2} x - 3 \left(0\right) = 0$
So we can conclude that $x = n \pi$ correspond to points of inflexion

When $x = n \pi - \frac{\pi}{2} \implies \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 \sin x \left(0\right) - 3 {\sin}^{3} \left(n \pi - \frac{\pi}{2}\right)$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 3 {\sin}^{3} \left(n \pi - \frac{\pi}{2}\right)$ .... [3]

Now $\sin \left(A - B\right) \equiv \sin A \cos B - \cos A \sin B$
$\therefore \sin \left(n \pi - \frac{\pi}{2}\right) = \sin n \pi \cos \left(\frac{\pi}{2}\right) - \cos n \pi \sin \left(\frac{\pi}{2}\right)$
$\therefore \sin \left(n \pi - \frac{\pi}{2}\right) = \left(\sin n \pi\right) \left(0\right) - \cos n \pi \left(1\right)$
$\therefore \sin \left(n \pi - \frac{\pi}{2}\right) = - \cos n \pi$

And so, substituting into [3] gives us:
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 3 {\left(- \cos n \pi\right)}^{3}$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 3 {\cos}^{3} n \pi$

So we can conclude that
$\left\{\begin{matrix}\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 < 0 & n \text{ odd" & => & "maximum" \\ (d^2y)/dx^2>0 & n " even" & => & "minimum}\end{matrix}\right.$

So Points of Inflection occur when $x = n \pi$,
ie $x = - 3 \pi , - 2 \pi , - \pi , 0 , \pi , 2 \pi , 3 \pi , \ldots$

Maximum points occur when $x = n \pi - \frac{\pi}{2}$ and n is odd
ie $x = \frac{- 11 \pi}{2} , \frac{- 7 \pi}{2} , \frac{- 3 \pi}{2} , \frac{\pi}{2} , \frac{5 \pi}{2} , \frac{9 \pi}{2} , \ldots$

Minimum points occur when $x = n \pi - \frac{\pi}{2}$ and n is even
$x = \frac{- 9 \pi}{2} , \frac{- 5 \pi}{2} , \left(- \frac{\pi}{2}\right) , \frac{3 \pi}{2} , \frac{7 \pi}{2} , \ldots$

This can be visualised by the graph