# Thread: Minimum and Maximum question

1. ## Minimum and Maximum question

I hope I am understanding the concept correctly. I'm just checking my work here...

$\displaystyle f(x)=2+3x^2-x^3$

$\displaystyle f'(x)=-3x^2+6x$ =$\displaystyle 3x(x-2)$?

So 0 and 2 are the critical numbers?

If this is the case how does the second derivative give me the maximum and minimum?

2. Originally Posted by XIII13Thirteen
I hope I am understanding the concept correctly. I'm just checking my work here...

$\displaystyle f(x)=2+3x^2-x^3$

$\displaystyle f'(x)=-3x^2+6x$ =$\displaystyle 3x(x-2)$?

So 0 and 2 are the critical numbers?

If this is the case how does the second derivative give me the maximum and minimum?
The second derivative gives a property of the curve known as "concavity." The archetypal function to understand concavity from is a parabola. If the parabola opens upward we may show that the concavity is constant and positive. If the parabola opens downward we may show that the concavity is constant and negative.

Your function has a second derivative of
$\displaystyle f"(x) = -6x + 6$

So at the critical point if $\displaystyle f"(x)$ is positive/negative then the point corresponds to a relative minimum/maximum.

-Dan