Aliena

2/13/2013 12:24:41 AM

hey try this...

The only possible inflexion points will happen where

(d²y)/( dx²)   = 0

From specified function as:

(dy)/(dx) = 3x² and (d²y)/( dx²)   = 6x

Equating the second derivative to zero, we include

6x = 0 or x = 0

We test whether the point at that x = 0 is an inflexion point as follows

While x is slightly less than 0, ((d²y)/(dx²)) < 0; it means a downward concavity

While x is slightly larger than 0, ((d²y)/(dx²)) > 0;  it means an upward concavity

Hence we have a point of inflexion at point x = 0 since the concavity of the curve changes as we pass from the left to the right of x = 0