What are the absolute extrema of # f(x)= x^3 -3x+1 in [0,3]#?

1 Answer
Feb 14, 2016

Answer:

Absolute minimum of #-1# at #x=1# and an absolute maximum of #19# at #x=3#.

Explanation:

There are two candidates for the absolute extrema of an interval. They are the endpoints of the interval (here, #0# and #3#) and the critical values of the function located within the interval.

The critical values can be found by finding the function's derivative and finding for which values of #x# it equals #0#.

We can use the power rule to find that the derivative of #f(x)=x^3-3x+1# is #f'(x)=3x^2-3#.

The critical values are when #3x^2-3=0#, which simplifies to be #x=+-1#. However, #x=-1# is not in the interval so the only valid critical value here is the one at #x=1#. We now know that the absolute extrema could occur at #x=0,x=1,# and #x=3#.

To determine which is which, plug them all into the original function.

#f(0)=1#
#f(1)=-1#
#f(3)=19#

From here we can see that there is an absolute minimum of #-1# at #x=1# and an absolute maximum of #19# at #x=3#.

Check the function's graph:

graph{x^3-3x+1 [-0.1, 3.1, -5, 20]}