#y = 2x^3 + 9x^2 - 24x - 1#

#(dy)/(dx) = 6x^2 + 18x - 24#

#(d^2y)/(dx^2) = 12x + 18#

For critical points need #(dy)/(dx) = 0#.

#therefore 0 = 6x^2 + 18x - 24#

Divide both sides by 6:

#x^2 + 3x - 4 = 0#

#(x+4)(x-1) = 0#

#implies x = -4 " or " x = 1#

For the nature of these points, need to examine the curvature which is given by the second derivative:

#color(red) (x=-4)#

#(d^2y)/(dx^2)|_(x=-4) = 12(-4) + 18 = -30 < 0#

This means that #x = -4# is a maximum.

#y(-4) = 2(-4)^3 + 9(-4)^2 - 24(-4)-1#

#y(-4) = -128 + 144 + 96 - 1 = 111#

#therefore (-4, 111) " is a maximum"#

#color(red)(x=1)#

#(d^2y)/(dx^2)|_(x=1) = 12(1) + 18 = 30 > 0#

This means that #x = 1# is a minimum.

#y(1) = 2(1)^3 + 9(1)^2 - 24(1)-1#

#y(1) = 2 + 9 - 24 - 1 = -14#

#therefore (1,-14) " is a minimum"#