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

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1 Answer
Jim H · Ken C.
May 25, 2016

On #[0,3]#, the maximum is #19# (at #x=3#) and the minimum is #-1# (at #x=1#).

Explanation:

To find the absolute extrema of a (continuous) function on a closed interval, we know that the extrema must occur at either crtical numers in the interval or at the endpoints of the interval.

#f(x) = x^3-3x+1# has derivative

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

#3x^2-3# is never undefined and #3x^2-3=0# at #x=+-1#.

Since #-1# is not in the interval #[0,3]#, we discard it.

The only critical number to consider is #1#.

#f(0) = 1#

#f(1) = -1# and

#f(3) = 19#.

So, the maximum is #19# (at #x=3#) and the minimum is #-1# (at #x=1#).

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