How do use the first derivative test to determine the local extrema #f(x) = x³+3x²-9x+15#?

1 Answer
Aug 9, 2015

Answer:

#x=-3# is a local maximum and #x=1# a local minimum.

Explanation:

First we look for points that might be the extrema by solving #f'(x)=0#:
#f'(x)=3x^2+6x-9=3(x+3)(x-1)#
#3(x+3)(x-1)=0 => x=-3 vv x=1#
#x=-3# and #x=1# are two potential extrema.

Now, we can use the graph of the first derivative to determine whether those points are extrema and which type of extrema:

if when passing through point #x# the first derivative changes it's sign from positive to negative then #x# is a local maximum
and if from negative to positive - a locac minimum.

graph{3(x+3)(x-1) [-10, 10, -5.21, 5.21]}

In this case #x=-3# is a local maximum and #x=1# a local minimum.