#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"#
