Question

What is the local maxima or minima of `x^3 - 3x^2 + 1`?

Answer 1

maximum at (0 , 1 )
minimum at (2 , -3 )

Explanation:

To find the x-coordinates of maxima / minima differentiate the

function and equate to zero ( the gradients of tangents at

stationary points = 0 ).

let y = `x^3 - 3x^2 + 1`

now `dy/dx = 3x^2 - 6x = 0`

hence 3x (x - 2 ) = 0 and x = 0 or x = 2

To find y-coordinates substitute these values into y .

x = 0 : y = 1 hence (0 , 1 ) is a stationary point.( S.P )

x = 2 : y = `(2)^3 - 3(2)^2 + 1 = - 3`

hence ( 2 , - 3 ) is an S.P

To find the 'Nature' of these points use the second derivative.

If `(d^2y)/dx^2` < 0 then maximum S.P

If `(d^2y)/dx^2`> 0 then minimum S.P

now `(d^2y)/dx^2 = 6x - 6`

`x = 0 :( d^2 y)/dx^2 = - 6 < 0 rArr (0 ,1 ) color(black)(" is a max")`

`x = 2 : (d^2y)/dx^2 = 12 - 6 = 6 > 0 rArr (2 , -3 ) color(black)( "is a min")`

graph{x^3-3x^2+1 [-10, 10, -5, 5]}