Question

What is the point(s) of inflection of the curve of `g(x) = -4x^3-2x^2+x - 1`?

Answer 1

There is a non-stationary point of inflection at `(-1/6, -65/54)`

Explanation:

We have:

`g(x) = -4x^3-2x^2+x-1`
graph{-4x^3-2x^2+x-1 [-5, 5, -5, 5]}

We would normally look for critical points, that is coordinates where `dy/dx=0`, however the question does not require this so it will be skipped.

The first derivative is then:

`g'(x) = -12x^2-4x+1`

So the second derivative is then:

`g''(x) = -24x-4`

We look for inflection points, which are coordinates where the second derivative vanishes:

`g''(x) = 0 => -24x-4 = 0`
`:. 24x=-4`
`:. x = -1/6`

A point of inflection is graded as a stationary point of inflection of the first derivative vanished at the point, otherwise as a non-stationary point of inflection

So when `x=-1/6`, we have:

`g(-1/6) = 4(1/216) - 2(1/36) -1/6 -1`
`" " = -65/54`

And:

`g'(-1/6) = -12(1/36) + 4(1/6) + 1`
`" " = 4/3`

Hence, There is a non-stationary point of inflection at `(-1/6, -65/54)`