If #f'(x) = (x-8)^9 (x-4)^7 (x+3)^7#, what are the local minima and maxima of #f(x)#?

1 Answer
Aug 19, 2016

Answer:

There is a local maximum at #x = 4# and two local minima at
#x = -3# and #x = 8#

Explanation:

#f'(x) = (x-8)^9 (x-4)^7 (x+3)^7# so the stationary points are

#x =-3, x = 4# and #x = 8#. At those points the behavior of #f(x)# is described by

#f(x) approx f(x_0) + C (x-x_0)^n# where #n# is defined by the firts non null derivative of #f(x)# at #x_0#, and #C# is #1/(n!)(d^nf(x_0))/(dx^n)#. In the present case we have:

1) For the stationary point #x=-3# we have #n = 8# and
#C = 1941871315289213/8 > 0# minimum

2) For the stationary point #x=4# we have #n = 8# and
#C = -26985857024 < 0# maximum

3) For the stationary point #x=8# we have #n = 10# and
#C = 159638904832/5 > 0# minimum

So there is a local maximum at #x = 4# and two local minima at
#x = -3# and #x = 8#