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
$\displaystyle f'_(x)=x^2+4x+3$
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 $\displaystyle f''_{(x_2)}<0$