#f(x) = -2/9x^3 -2/3x^2+16/3x+160/9#

First, let's clean the expression up by factoring out #-2/9#.

#f(x) = -2/9(x^3 + 3x^2 - 24x - 80)#.

Now, we'll find the critical numbers by finding the numbers in the domain at which the derivative is #0# or fails to exist.

The domain of #f# is all real numbers.

#f'(x) = -2/9(3x^2+6x-24)#.

The derivative exists for all real #x#, so now we need to solve #f'(x)=0# to find the critical numbers.

#-2/9(3x^2+6x-24)=0# is equivalent to

#-2/3(x^2+2x-8)=0#

#-2/3(x+4)(x-2) = 0#

#x=-4# #" "# or #" "# #x=2#.

These are the critical numbers for #f#. They determine three intervals on the number line.

We look at the sign of #f'# on each interval to determine whether #f# is increasing or decreasing on the interval

#{: (bb "Interval", bb"Sign of "f',bb" Incr/Decr"),
((-oo,-4)," " -" ", " "" Decr"),
((-4,2), " " +, " " " Incr"),
((2 ,oo), " " -, " "" Decr")
:}#

#f# has a local minimum of #0# at #-4#, and

#f# has a local maximum of 24 at #2#.

**Calculations**

#f(-4) = -2/9[(-4)^3+3(-4)^2-24(-4)-80]#

# = -2/9[-4(16)+3(16)+6(16)-5(16)]#

# = -2/9[0(16)] = 0#

and

#f(2) = -2/9[(2)^3+3(2)^2-24(2)-80]#

# = -2/9[2(4)+3(4)-12(4)-20(4)]#

# = -2/9[-27(4)] = (2cancel((27))^3(4))/cancel(9) = 24#