Question

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

Answer 1

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(x)`, we follow the usual procedure and set `f'(x)=0`.
`f(x)=12x^5+45x^4-80x^3+6` so
`f'(x)=60x^4+180x^3-240x^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'(x)=60x^2(x^2+3x-4)=60x^2(x+4)(x-1)`.

So `f'(x)=0rArr x=`-4, 0, or 1.

To categorise these consider the second derivative:
`f''(x)=240x^3+540x^2-480x=60x(4x^2+9x-8)`

`f''(-4)=-4800<0`
`f''(0)=0`
`f''(1)=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'''(x)=720x^2+1080x-480`, so `f'''(0)=-480!=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]}