Question

How do you find the minimum values for `f(x)=2x^3-9x+5` for `x>=0`?

Answer 1

You first take the derivative

`f(x)=2x^3-9x+5->f'(x)=6x^2-9`

You have an extreme if `f'(x)=0`

`6x^2-9=0->6x^2=9->x^2=9/6=1.5`

`x=+-root 2 1.5=+-1.225...`

And we keep only the positive value.
We put that in the formula to get `y`

Answer : `x~~1.225` and `y~~-2.358`
graph{2x^3-9x+5 [-25.66, 25.68, -12.83, 12.81]}