1.For the following function:

y= x^3/3+ 2x^2 + 3x - 1

a.Find the stationary points.

b.Use the second-order derivatives to test whether the stationary points are at a

maximum or minimum value of.

thank you for your help

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- Jul 19th 2010, 02:23 PMarslanstationary points?
**1.****For the following function:**

**y= x^3/3+ 2x^2 + 3x - 1**

**a.****Find the stationary points.**

**b.****Use the second-order derivatives to test whether the stationary points are at a**

**maximum or minimum value of****.**

**thank you for your help**

- Jul 19th 2010, 02:46 PMAlso sprach Zarathustra
a. find f''(x), then x1 and x2 of f'(x)=0

b. find f''(x), if f''(xt)>0 then xt min. if f''(xt),0 xt max. if f''(xt)=0 and f'''(xt)!=0 then it is nor max/min. - Jul 20th 2010, 05:08 AMarslan
thanks however i am clueless to finding f''(x), then x1 and x2 of f'(x)=0

- Jul 20th 2010, 07:10 AMyeKciM
$\displaystyle f'_(x)=x^2+4x+3$

when u put $\displaystyle f'(x)=0$ means that $\displaystyle x^2+4x+3=0$

$\displaystyle x_1=-1 , x_2=-3$ There u have stationary points...

$\displaystyle f''(x)=(f'(x))'=2x+4$

lol... u can go from here ?

just put stationary points in $\displaystyle f''_{(x)}$

$\displaystyle f''_{(x_1)}=2 $ so it's min because it's $\displaystyle f''_{(x_1)}>0$

$\displaystyle f''_{(x_2)}=-2}$ so it's max because it's $\displaystyle f''_{(x_2)}<0$