local minimum of function

f(x)=2x^3-30x^2+96x+9

Take the derivative and set to 0

f ' (x)  =  6x^2 - 60x + 96  = 0      divide everything by 6

x^2 - 10x + 16  = 0     factor

(x - 8) ( x- 2) = 0     so...the critical points are x = 8 and x = 2

Take the second derivative

f '' (x)  =  2x - 10

When x = 8,   f " (x) > 0   so this is a minimum at (8, -119)

When x = 2,   f " (x) < 0    so this is a max at (2, 97)

Here's a graph :   https://www.desmos.com/calculator/w1v9hktjpi

find the local minimum of f(x)=2x^3-30x^2+96x+9

min{2 x^3-30 x^2+96 x+9} = -119 at x = 8