Question

How do you find points of inflection for `(x) = 2x(x-4)^3`?

Answer 1

There are two points of inflection : one when `x_1=2` and another when `x_2=4`.

The points of inflection of a function are given by the study of the sign the second derivative. Those are the points where the concavity of a function changes.

Let's derive `f(x) = 2x(x-4)^3` :

`f'(x) = (2x)' * (x-4)^3 + ((x-4)^(3))' * (2x)`

`f'(x) = 2(x-4)^3 + 6x(x-4)^2 = (x-4)^2(2x-8+6x)`

`f'(x) = 8(x-4)^2(x-1)`

Now, let's derive `f'(x)` :

`f''(x) = ((x-4)^2)' * 8 * (x-1) + (x-1)' * 8 * (x-4)^2`

`f''(x) = 16(x-4)(x-1) + 8(x-4)^2`

`f''(x) = (x-4)(16x-16+8x-32)`

`f''(x) = (x-4)(24x-48) = 24(x-4)(x-2)`

`f''(x) = 0`, when `x_1=2` and `x_2=4`.

Therefore, your function has two points of inflection : one when `x_1=2` and another when `x_2=4`.

That's it!