# Examine the following function for maxima and minima using the second derivative method: y=x^3-3x^2-9x+5

llltkl | Student

Given, `y=x^3-3x^2-9x+5`

Differentiating y with respect to x,

`y'=3x^2-3*2x-9=3x^2-6x-9`

At local maximum or local minimum, `y'=0`

So, `3x^2-6x-9=0`

`rArr x^2-2x-3=0`

`rArr x^2-3x+x-3=0`

`rArr (x-3)(x+1)=0`

`x=3, -1`

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

`y''=6x-6`

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.

`y(3)=3^3-3*3^2-9*3+5=-22`

and `y(-1)=(-1)^3-3(-1)^2-9(-1)+5=10`

Therefore, the minimum point is at (3,-22) and the maximum point is at (-1,10).