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

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Answer 1 · 2015-02-05T13:43:34

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]}

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