Question

How do you graph `y = 2x^3 - 4x + 1` using the first and second derivative?

Answer 1

You can start by using the first derivative to find points of maximum and minimum:
`y'=6x^2-4`
You set the first derivative equal to zero:
`6x^2-4=0` which gives you:
`x=+-sqrt(2/3)`

So:
maximum: `x=-sqrt(2/3)` and `y=3.17`
minimum: `x=sqrt(2/3)` and `y=-1.17`

You use the second derivative to find inflection points:
`y''=12x`
Setting it equal to zero you get the point of inflection:
`x=0` `y=1`

And finally the graph: