# How do you find the inflection points of f(x)= 12x^5+45x^4-80x^3+6?

Jun 22, 2018

Maximum: x=-4
Inflection: x=0
Minimum: x=1

#### Explanation:

In general, this is a hard problem for a quintic equation. But this particular quintic has some missing low order terms that help us a great deal.

To find the inflection and turning points of $f \left(x\right)$, we follow the usual procedure and set $f ' \left(x\right) = 0$.
$f \left(x\right) = 12 {x}^{5} + 45 {x}^{4} - 80 {x}^{3} + 6$ so
$f ' \left(x\right) = 60 {x}^{4} + 180 {x}^{3} - 240 {x}^{2}$

Now this is a quartic equation. A full "solution by radicals" of the quartic is known, but it is long and complex (and painful to perform!). Fortunately, we can factorise this equation:

$f ' \left(x\right) = 60 {x}^{2} \left({x}^{2} + 3 x - 4\right) = 60 {x}^{2} \left(x + 4\right) \left(x - 1\right)$.

So $f ' \left(x\right) = 0 \Rightarrow x =$-4, 0, or 1.

To categorise these consider the second derivative:
$f ' ' \left(x\right) = 240 {x}^{3} + 540 {x}^{2} - 480 x = 60 x \left(4 {x}^{2} + 9 x - 8\right)$

$f ' ' \left(- 4\right) = - 4800 < 0$
$f ' ' \left(0\right) = 0$
$f ' ' \left(1\right) = 300 > 0$

So -4 is a maximum of the function and +1 is a minimum. To categorise 0 we need to take another derivative.

$f ' ' ' \left(x\right) = 720 {x}^{2} + 1080 x - 480$, so $f ' ' ' \left(0\right) = - 480 \ne 0$

Thus 0 is a point of inflection.

We want to sanity check our answer by comparing to the graph of the function, but the maximum at -4 is off the scale compared to the other points of interest. Plot it twice, once zoomed in, once zoomed out:
graph{12x^5+45x^4-80x^3+6 [-5, 2, -20, 20]}
graph{12x^5+45x^4-80x^3+6 [-5, 2, -1000, 5000]}