Question

If `f^ { \prime } ( x ) = - 24x ^ { 3} + 9x ^ { 2} + 3x + 1`, and there are two points of inflexion on the graph of `f`, what are the x-coordinates of these points?

Answer 1

Points of inflexion are `x=0.36436` and `-0.11436`

Explanation:

The ponts of inflexion are those points at which the value of second derivative iszero. These are points at which function changes from increasing to decreasing or vice versa.

As `f'(x)=-24x^3+9x^2+3x+1`

`f''(x)=-72x^2+18x+3`

and when `-72x^2+18x+3=0` we have `24x^2-6x-1=0`

i.e. `x=(6+-sqrt(6^2-4*24*(-1)))/48`

= `(6+-sqrt132)/48=1/8+-sqrt33/24`

i.e. `x=0.36436` or `-0.11436`

Hence these are points of inflexion,

Observe that `f(x)=-6x^4+3x^3+1.5x^2+x` and its graph is

graph{-6x^4+3x^3+1.5x^2+x [-2.5, 2.5, -1.25, 1.25]}