# If y = 3x^5 - 5x^3, what are the points of inflection of the graph f (x)?

Oct 12, 2016

$x = 0$

#### Explanation:

$y = 3 {x}^{5} - 5 {x}^{3}$
$y ' = 15 {x}^{4} - 15 {x}^{2}$
At a critical point $y ' = 0 \implies 15 {x}^{4} - 15 {x}^{2} = 0$
$\therefore 15 {x}^{2} \left({x}^{2} - 1\right) = 0$
$\therefore 15 {x}^{2} \left(x - 1\right) \left(x + 1\right) = 0$
So there are three critical points, when $x = 0 , x = \pm 1$

To determine the nature of these critical points we look at the second derivative:
$y ' = 15 {x}^{4} - 15 {x}^{2}$
$\therefore y ' ' = 60 {x}^{3} - 30 x$
$\therefore y ' ' = 30 x \left(2 {x}^{2} - 1\right)$

So When:
$x = 0 \implies y ' ' = 0$
$x = - 1 \implies y ' ' = \left(- 30\right) \left(2 - 1\right) = - 30$
$x = 1 \implies y ' ' = \left(30\right) \left(2 - 1\right) = 30$

So the nature of the critical points is as follows
$x = 0$ point oi inflection
$x = - 1$ maximum
$x = 1$ minimum

Oct 12, 2016

The points of inflection (with rationalized denominators) are: $\left(\frac{- \sqrt{2}}{2} , \frac{7 \sqrt{2}}{8}\right)$, $\left(0 , 0\right)$ $\left(\frac{\sqrt{2}}{2} , \frac{- 7 \sqrt{2}}{8}\right)$

#### Explanation:

Points of inflection are points on the graph at which the concavity (and the sign of the second derivative) change.

$y = 3 {x}^{5} - 5 {x}^{3}$

$y ' = 15 {x}^{4} - 15 {x}^{2}$

$y ' ' = 60 {x}^{3} - 30 x$

The zeros of $y ' '$ are found by solving

$y ' ' = 30 x \left(2 {x}^{2} - 1\right) = 0$

The solutions are $x = - \frac{1}{\sqrt{2}}$, $0$, and $\frac{1}{\sqrt{2}}$.

Each of these is a zero of the polynomial $y ' '$ with multiplicity $1$, so the sign changes at each of them. (Or make a sign table, chart, diagram, whatever you've been taught to call it.)

Therefore each is the $x$-value of a point of inflection. To find the points, find the corresponding $y$-values.

The points of inflection (with rationalized denominators) are: $\left(\frac{- \sqrt{2}}{2} , \frac{7 \sqrt{2}}{8}\right)$, $\left(0 , 0\right)$ $\left(\frac{\sqrt{2}}{2} , \frac{- 7 \sqrt{2}}{8}\right)$