Question

What are the points of inflection of `f(x)=3x^5 - 5x^3`#?

Answer 1

They are
`(-sqrt2/2, sqrt2/4) , (0, 0) , (1, -sqrt2/4)`

Explanation:

Possible point of inflection is determine by setting the derivative equal to zero. Possible point on inflection happen when there is changes in concavity (refer to the graph below)

`f(x) = 3x^5 -5x^3`

`f'(x) = 15x^4 -15x^2`

`f''(x) = 60x^3 -30x`

Set the second derivative equal to zero

`60x^3 - 30x = 0`

`30x(2x^2 -1) = 0`

`30x= 0 => x = 0`

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

`x^2 =1/2>x = +-sqrt(1/2) => +- sqrt(2)/2`

Coordinates point for P.O.I (Point of inflection)

`f(-sqrt2/2) = 3(-sqrt2/2)^5 - 5(-sqrt2/2)`

`(3)(-sqrt2/8) -5(-sqrt2/8)=-3sqrt2/8+5sqrt2/8`

`2sqrt2/8=>sqrt2/4==> (-sqrt2/2,sqrt2/4)`

`f(0) = 0-0 =0 => (0, 0)`

`f(sqrt2/2) = 3(sqrt2/2)^5 - 5(sqrt2/2)`

`(3)(sqrt2/8) -5(sqrt2/8)=3sqrt2/8-5sqrt2/8`

`-2sqrt2/8=>-sqrt2==> (-sqrt2/2,-sqrt2/4)`

graph{3x^5-5x^3 [-10, 10, -5, 5]}