Question

What are the points of inflection, if any, of `f(x) = x^6 + 3x^5 - (15/2)x^4 - 40x^3 - 60x^2 + 8x + 5`?

Answer 1

Points of inflection occur at `x=-1,-2` and `2`.

Explanation:

At point of inflection `(d^2f(x))/(dx^2)=0`

As `f(x) = x^6+3x^5-(15/2)x^4-40x^3-60x^2+8x+5`

`(df)/(dx)=6x^5+15x^4-30x^3-120x^2-120x+8`

and `(d^2f(x))/(dx^2)=30x^4+60x^3-90x^2-240x-120`

Hence points of inflection are given by `30x^4+60x^3-90x^2-240x-120=0`

or `x^4+2x^3-3x^2-8x-4=0`

Using factor theorem we have zeros at `-1,-2` and `2`

`(x-2)(x+2)(x+1)^2=0`

Hence, points of inflection occur at `x=-1,-2` and `2`.

Graph not drrawn to scale (shrunk vertically).
graph{x^6+3x^5-(15/2)x^4-40x^3-60x^2+8x+5 [-4, 4, -200, 200]}