# stationary points?

• Jul 19th 2010, 02:23 PM
arslan
stationary 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.

• Jul 19th 2010, 02:46 PM
Also 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 AM
arslan
thanks however i am clueless to finding f''(x), then x1 and x2 of f'(x)=0
• Jul 20th 2010, 07:10 AM
yeKciM
\$\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\$