How do you locate the absolute extrema of the function #f(x)=x^3-12x# on the closed interval [0,4]?

1 Answer
Jul 1, 2015

Answer:

The absolute extrema of a function, #f#, that is continuous on a closed interval must occur at either a critical number for #f# in the interval, or at an endpoint of the interval.

Explanation:

So consider:

#f(x) =x^3-12x#

#f'(x) = 3x^2-12#.

#f'# is polynomial, so every critical number for #f# is a zero for #f'#.

The zeros of #f'# are: #3(x^2-4)=0#, #x=-2, 2#. But #-2# is not in the interval #[0,4]#.
So the only critical number in the interval is #2#.

The minimum and maximum must occur at one of the values

#x=# #0# or #2# or #4#.

To finish, evaluate #f# at each of these values.

#f(0) = 0#

#f(2) = 8-24 = -16#

#f(4) = 64-48 = 16#

The minimum is #-16# (it occurs at #2#)
The maximum is #16# (it occurs at #4#).

To understand what we've done it may help to see the graph:

graph{y=(x^3-12x)*sqrt(4-(x-2)^2)/(sqrt(4-(x-2)^2)) [-52.2, 40.28, -23.87, 22.4]}