Question

How do you find the points of Inflection of `f(x)=2x(x-4)^3`?

Answer 1

Below

Explanation:

`f(x)=2x(x-4)^3`
`f'(x)=2xtimes3(x-4)^2+2(x-4)^3`
`f'(x)=2(x-4)^2(3x+x-4)=2(x-4)^2(4x-4)`
`f''(x)=2times(2(x-4)(4x-4)+(x-4)^2(4))`
`f''(x)=2(8x^2-40x+32+4x^2-32x+64)`
`f''(x)=2(12x^2-72x+96)`
`f''(x)=2(12)(x^2-6x+8)`
`f''(x)=24(x-4)(x-2)`

For points of inflexion, `f''(x)=0`

ie
`24(x-4)(x-2)=0`
`x=4,2`

Test `x=4`
At `x=3`, `f''(x)=-24`
At `x=4`, `f''(x)=0`
At `x=5`, `f''(x)=72`

Therefore, there is a change in concavity so there is a point of inflexion at x=4

Test `x=2`
At `x=1`, `f''(x)=72`
At `x=2`, `f''(x)=0`
At `x=3`, `f''(x)=-24`

Therefore there is a change in concavity so there is a point of inflexion at x=2

Answer 2

Point of inflection is at `x=4` or `x=8/3`

Explanation:

Points of Inflection appear where the curve changes from being concave to convex or vice versa. This happens when second derivative i.e. `(d^2f)/(dx^2)=0`

As `f(x)=2x(x-4)^3`

`(df)/(dx)=2(x-4)^3+2x(x-4)^2=2x^3-24x^2+96x-128+2x^3-16x^2+32x=4x^3-40x^2+128x-128`

and `(d^2f)/(dx^2)=12x^2-80x+128`

This is `0`, when `12x^2-80x+128=0`

or `3x^2-20x+32=0`

or `(x-4)(3x-8)=0`

Hence point of inflection is at `x=4` and `x=8/3`

Graph not drawn to scale.

graph{2x(x-4)^3 [-10, 10, -70, 30]}