Question

Find the points of inflection: `y=x^(1/3)(x-4)`?

Answer 1

`(x,y)=(0.712,-2.396)`happens to be the point of inflection where the curvature is zero

Explanation:

`y`=#x^(1/3) (x-4)#
Points of inflection are the points where there is a change in curvature of the function.
This happens when the curvature is zero.
given:
y=#x^(1/3) (x-4)#
Cubing both sides
`y^3`=`x(x-4)^3`

Differentiating wrt x
`3y^2y'`=`x(3(x-4)^2)+(x-4)^3`
=`(x-4)^2(3x+x-4)`
=`(x-4)^2(4x-4)`
=`(x-4)^2 4(x-1)`

Thus,
`3y^2y'`=`4(x-1)(x-4)^2`
Differentiating again wrt x
`3(y^2y''+y'(2y))`=`4((x-1)2(x-4)+(x-4)^2)`
=`4(x-4)(2(x-1)+(x-4))`
=`4(x-4)(2x-2+x-4)`
=`4(x-4)(3x-6)`
=`4(x-4)3(x-2)`
=`12(x-4)(x-2)`
`3(y^2y''+y'(2y))`=`12(x-4)(x-2)`

Dividing by 3 on both sides
`y^2y''+2yy'`=`4(x-4)(x-2)`
`y''=0`implies
`2yy'`=`4(x-2)(x-4)`
`yy'`=`2(x-4)(x-2)`

Rewriting the equation obtained after first differentiation
`3y^2y'`=`4(x-1)(x-4)^2`
Rearranging,
`3y.yy'`=`4(x-1)(x-4)^2`
Substituting for `yy'`
`3y(2(x-4)(x-2))`=`4(x-1)(x-4)^2`
Implies
`6y(x-4)(x-2)`=`4(x-1)(x-4)^2`
Simplifying,
`y(x-2)`=`4(x-1)(x-4)`
`y`=`4(((x-1)(x-4))/(x-2))`
Substituting for y
#x^(1/3) (x-4)#=`4(((x-1)(x-4))/(x-2))`
Further,
#x^(1/3) #=`4((x-1)/(x-2))`
Cubing both sides,
`x`=`64(x-1)^3 /(x-2)^3`

Cross multiplying
`x(x-2)^3`=`64(x-1)^3`
Expanding the polynomials
`x(x^3-3*2x^2+3.2^2*x-2^3)`=`64(x^3-3x^2+3x-1)`
`x^4-6x^3+12x^2-8x`=`64x^3-192x^2+192x-64`

Transposing the terms on R.H.S. to L.H.S
`x^4-6x^3+12x^2-8x-(64x^3-192x^2+192x-64)`=`0`
`x^4-6x^3+12x^2-8x-64x^3+192x^2-192x+64`=`0`

The root is investigated by trial and error and found to be x=0.712 correct to three decimal places
Substituting x = 0.712, y will be
`y`=#x^(1/3) (x-4)#
=`0.712^(1/3) (0.712-4)`
=`-2.396`
Hence, `(x,y)=(0.712,-2.396)`happens to be the point of inflection