The Second Derivative

The second differentiail criterion al 
\[\]
 lows us to classify turning points of a function  
\[f(x)\]
 (points for which  
\[\frac{df}{dx}=0\]
).
Suppose there is a function  
\[f(x)\]
  and that at some point  
\[x_0\]
,  
\[\frac{df}{dx}=0\]
.
Then  
\[(x_0, f(x_0))\]
  is a turning point of the function  
\[f(x)\]
  and
If  
\[\frac{d^2f}{dx^2} < 0\]
  then  
\[(x_0, f(x_0))\]
  is at a maximum for the function  
\[f(x)\]
.
If  
\[\frac{d^2f}{dx^2} > 0\]
  then  
\[(x_0, f(x_0))\]
  is at a minimum for the function  
\[f(x)\]
.
Example: Let  
\[f(x)=x^3-4x^2-3x\]
. Then  
\[\frac{df}{dx}=3x^2-8x-3\]
. Ti find the turning point set  
\[\frac{df}{dx}=3x^2-8x-3=(3x+1)(x-3)=0\]
.
Then the turning points are  
\[x_1 =- \frac{1}{3}, \: x_2 =3\]
.
\[\frac{d^2f}{dx^2}=6x-8\]
.
When  
\[x=- \frac{1}{3}, \: \frac{d^2f}{dx^2}=6 \times - \frac{1}{3}-8=-10 \lt 0\]
.
Hence  
\[x_1= - \frac{1}{3}\]
  is at a maximum.
When  
\[x=3, \: \frac{d^2f}{dx^2}=6 \times 3-8=10 \gt 0\]
.
Hence  
\[x_1= - \frac{1}{3}\]
  is at a minimum.

second derivative criterion