Question

How do you find the inflection points of the graph of the function: `f(x) = 2x(x-4)^3`?

Answer 1

The inflection points are given when the second derivative is equaled to zero, because the second derivative indicates the change in concavity of the function.

`(dy)/(dx)=2(x-4)^3+2x(3(x-4)^2)=2(x-4)^3+6x(x-4)^2`

`(dy)/(dx)=[(2x-8)+(6x)][(x-4)^2]=color(green)((8x-8)(x-4)^2)`

`(d^2y)/(dx^2)=8(x-4)^2+(8x-8)(2x-8)`

Developing:

`(d^2y)/(dx^2)=8(x^2-8x+16)+(16x^2-80x+64)=24x^2-144x+192`

Equaling the second derivative to zero:

`24(x^2-6x+8)=0`

`(6+-sqrt(36-(4)(1)(8)))/2`
`(6+-sqrt(4))/2`
`(6+-2)/2`

`x_1=4` and `x_2=2`

Substituting these values found in the second derivative there in the original function, we'll get the inflection points.

`f(4)=2(4)(4-4)^3=0`
`f(2)=2(2)(2-4)^3=4*(-8)=-32`

Thus, the inflection points are `A(4,0)` and `B(2,-32)`