Question

If `y = 3x^5 - 5x^3`, what are the points of inflection of the graph f (x)?

Answer 1

`x=0`

Explanation:

`y=3x^5-5x^3`
`y'=15x^4-15x^2`
At a critical point `y'=0=>15x^4-15x^2=0`
`:. 15x^2(x^2-1)=0`
`:. 15x^2(x-1)(x+1)=0`
So there are three critical points, when `x=0, x=+-1`

To determine the nature of these critical points we look at the second derivative:
`y'=15x^4-15x^2`
`:. y''=60x^3-30x`
`:. y''=30x(2x^2-1)`

So When:
`x=0 => y''=0`
`x=-1 => y''=(-30)(2-1)=-30`
`x=1 => y''=(30)(2-1)=30`

So the nature of the critical points is as follows
`x=0` point oi inflection
`x=-1` maximum
`x=1` minimum

Answer 2

The points of inflection (with rationalized denominators) are: `((-sqrt2)/2,(7sqrt2)/8)`, `(0,0)` `(sqrt2/2,(-7sqrt2)/8)`

Explanation:

Points of inflection are points on the graph at which the concavity (and the sign of the second derivative) change.

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

`y' = 15x^4-15x^2`

`y'' = 60x^3-30x`

The zeros of `y''` are found by solving

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

The solutions are `x=-1/sqrt2`, `0`, and `1/sqrt2`.

Each of these is a zero of the polynomial `y''` with multiplicity `1`, so the sign changes at each of them. (Or make a sign table, chart, diagram, whatever you've been taught to call it.)

Therefore each is the `x`-value of a point of inflection. To find the points, find the corresponding `y`-values.

The points of inflection (with rationalized denominators) are: `((-sqrt2)/2,(7sqrt2)/8)`, `(0,0)` `(sqrt2/2,(-7sqrt2)/8)`