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]}