Question

How do I find concavity and points of inflection for `y = 3x^5 - 5x^3`?

Answer 1

At `(0, 0)` There is point of inflection
At `(1,-2)` The curve is concave upwards.
At `(-1,2)` The curve is concave downwards.

Explanation:

Given -

`y=3x^5-5x^3`

Find the first derivative -

`dy/dx=15x^4-15x^2`
`dy/dx =0 =>15x^4-15x^2=0`

Then -

`15x^2(x^2-1)=0`
`15x^2=0`
`x=0`

`x^2-1=0`
`x^2=1`
`x=+-1`
`x=1`
`x=-1`

Find the second derivative -

`(d^2x)/(dx^2)=60x^3-30x`

At `x=0`

`(d^2y)/(dx^2)=60(0)^3-30(0)=0`

The value of the function -

`y=3(0)^5-5(0)^3=0`

At `(0, 0)`

`dy/dx=0; (d^2y)/(dx^2)=0`

Hence there is a point of inflection at `(0, 0)`

At `x=1`

`(d^2y)/(dx^2)=60(1)^3-30(1)=60-30=30>0`

The value of the function -

`y=3(1)^5-5(1)^3=03-5=-2`

At `(1, -2)`

`dy/dx=0; (d^2y)/(dx^2)>0`

Hence there is a minimum at `(1, -2)`
The curve is concave upwards

At `x=-1`

`(d^2y)/(dx^2)=60(-1)^3-30(-1)=-60+30=-30<0`

The value of the function -

`y=3(-1)^5-5(-1)^3=-3+5=2`

At `(1, 2)`

`dy/dx=0; (d^2y)/(dx^2)<0`

Hence there is a Maximum at `(1, 2)`
The curve is concave downwards.