What are the points of inflection, if any, of $f(x) =12x^5+15x^4–240x^3+6$ ?

Question

3377 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-24T11:47:34

The points of inflection are at:
$x=0,(-3+-sqrt(393))/8$

The points of inflection are given by when $f''(x)=0$ .

$f(x)=12x^5+15x^4-240x^3+6$
$f'(x) =60x^4+60x^3-720x^2$
$f''(x)=240x^3+180x^2-1440x$

$x=60x(4x^2+3x-24)$
(1)
$60x=0$
$x=0$

(2) and (3)
$4x^2+3x-24=0$
$x=(-3+-sqrt(393))/8$

What are the points of inflection, if any, of f(x) =12x^5+15x^4–240x^3+6f(x) =12x^5+15x^4–240x^3+6?