Examine the following function for maxima and minima using the second derivative method: y=x^3-3x^2-9x+5
Differentiating y with respect to x,
At local maximum or local minimum, `y'=0`
These two are the stationary points.
Now, at the maximum, y' or the slope of the graph must not rise further, so d/dx of y', i.e. y'' should be negative. Similarly, at the minimum, y' or the slope of the graph must not decline further, it must rise. So d/dx of y', i.e. y'' should be positive.
Differentiating y' with respect to x again
Examining the stationary points,
At `y''(3)=6*3-6=12` = (+)ve, so y must be minimum here.
At `y''(-1)=6*(-1)-6=-12` = (-)ve, so y must be maximum here.
Therefore, the minimum point is at (3,-22) and the maximum point is at (-1,10).