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

Feb 5, 2015

You first take the derivative

$f \left(x\right) = 2 {x}^{3} - 9 x + 5 \to f ' \left(x\right) = 6 {x}^{2} - 9$

You have an extreme if $f ' \left(x\right) = 0$

$6 {x}^{2} - 9 = 0 \to 6 {x}^{2} = 9 \to {x}^{2} = \frac{9}{6} = 1.5$

$x = \pm \sqrt[2]{1.5} = \pm 1.225 \ldots$

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

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