How do you find all points of inflection given $y=-x^5+2x^3+4$ ?

Question

1732 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-21T21:11:07

Inflection points at $x = 0$ and $x = +-sqrt(5/3)$

Find the second derivative:

$y' = -5x^4 + 6x^2$

$y'' = -20x^3 + 12x$

This will have points of inflection when $y'' =0$ .

$0 = -20x^3 + 12x$

$0 = -4x(5x^2 - 3)$

This means that $x = 0$ or $x = +-sqrt(5/3)$ .

Hopefully this helps!

How do you find all points of inflection given y=-x^5+2x^3+4y=-x^5+2x^3+4?