The function is
#f(x)=2x^3-3x^2-36x-7#
To fd the interval of increasing and decreasing, calculate the first derivative
#f'(x)=6x^2-6x-36#
To find the critical points, let #f'(x)=0#
#6x^2-6x-36=0#
#=>#, #x^2-x-6=0#
#=>#, #(x-3)(x+2)=0#
The critical points are
#{(x=3),(x=-2):}#
Build a variation chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##3##color(white)(aaaa)##+oo#
#color(white)(aaaa)##f'(x)##color(white)(aaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##↗##color(white)(aaaa)##↘##color(white)(aaaa)##↗#
The intervals of increasing are #x in (-oo,-2)uu(3,+oo)# and the interval of decreasing is #x in (-2,3)#
Calculate the second derivative
#f''(x)=12x-6#
The point of inflection is when #f''(x)=0#
#=>#, #12x-6=0#
#=>#, #x=1/2#
The intervals to consider are #(-oo,1/2)# and #(1/2,+oo)#
Build a variation chart
#color(white)(aaaa)##" Interval "##color(white)(aaaa)##(-oo,1/2)##color(white)(aaaa)##(1/2,+oo)#
#color(white)(aaaa)##" sign f''(x) "##color(white)(aaaaaa)##(-)##color(white)(aaaaaaaaaa)##(+)#
#color(white)(aaaa)##" f(x) "##color(white)(aaaaaaaaaaa)##nn##color(white)(aaaaaaaaaaaa)##uu#
The function is concave down in the interval #(-oo,1/2)# and concave down in the interval #(1/2,+oo)#.
graph{2x^3-3x^2-36x-7 [-26.64, 46.44, 1.46, 38]}
