Question

What are the points of inflection, if any, of `f(x)=2x(2x-3)^3`?

Answer 1

`(3/2,0)` is the only point of inflection

Explanation:

`f(x) = 2x(2x-3)^3`
graph{2x(2x-3)^3 [-2, 5, -11, 11]}

Using the product rule and chain rule we have;
`f'(x) = (2x)3(2x-3)^2(2) + (2)(2x-3)^3`
`:. f'(x) = 12x(2x-3)^2 + 2(2x-3)^3`
`:. f'(x) = (2x-3)^2(12x + 2(2x-3))`
`:. f'(x) = (2x-3)^2(12x + 4x - 6))`
`:. f'(x) = (2x-3)^2(16x - 6)`
`:. f'(x) = 2(2x-3)^2(8x - 3)` .... [1]

At critical points (min/max/poi) then `f'(x) = 0`
`f'(x) = 0 -> 2(2x-3)^2(8x - 3) = 0`
`:. x=3/8,3/2`

To determine the nature of these points we need to examine the second derivative to see if f'(x) is increasing (min), decreasing (max) or 0 (inflection):

Differentiating [1] wrt `x` we get;
`f''(x) = (2(2x-3)^2)(8) + (2(2)(2x-3)(2))(8x - 3)`
`:. f''(x) = 16(2x-3)^2 + 8(2x-3)(8x - 3)`

When `x=3/8 => f''(3/8) = 16(6/8-3)^2 > 0` (ie min)
When `x=3/2 => f''(3/2) = 0` (ie poi)

When `x=3/2 => f(3/2) = 0`

Hence `(3/2,0)` is the only point of inflection