Find the coordinates and determine the nature of the stationary points of curve y=(2/3)x^3+2x^2-6x+3
- Stationary points occur when dy/dx=0, therefore determine dy/dx first:
dy/dx= 2x2 + 4x - 6
2) solve dy/dx=0 for two values of x (using quadratic formula, if necessary):
(x+3)(x-1)=0 --> x= 1, -3
3) to determine the nature of the stationary points, second derivative is needed:
d2y/dx2= 4x + 4
4) substitute x-coordinates of stationary points from step 2 into secondary derivative to determine their nature:
x= 1 --> d2y/dx2= 8 > 0 therefore relative minimum
x= -3 --> d2y/dx2= -8 < 0 therefore relative maximum
5) substitute x-coordinates of stationary points from step 2 into y to get full coordinates:
y(1) = -1/3 --> (1, -1/3) is a rel. min.
y(-3) = 21 --> (-3, 21) is a rel. max.
BV
