What are the local extrema, if any, of #f (x) =x^3-3x+6#?

1 Answer
Nov 11, 2015

Answer:

#x^3-3x+6# has local extrema at #x=-1# and #x=1#

Explanation:

The local extrema of a function occur at points where the first derivative of the function is #0# and the sign of the first derivative changes.
That is, for #x# where #f'(x) = 0# and either #f'(x-varepsilon) <= 0 and f'(x+varepsilon) >= 0# (local minimum) or
#f'(x-varepsilon) >= 0 and f'(x+varepsilon) <= 0# (local maximum)

To find the local extrema, then, we need to find the points where #f'(x) = 0#.

#f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x+1)(x-1)#
so
#f'(x) = 0 <=> 3(x+1)(x-1) = 0 <=> x=+-1#

Looking at the sign of #f'# we get
#{(f'(x) > 0 if x < -1), (f'(x) < 0 if -1 < x < 1), (f'(x) > 0 if x > 1):}#

So the sign of #f'# changes at each of #x = -1# and #x = 1# meaning there is a local extremum at both points.

Note: From the change in signs, we can further tell that there is a local maximum at #x = -1# and a local minimum at #x = 1#.