What are the local extrema, if any, of #f (x) = x^3 - 6x^2 - 15x + 11 #?

1 Answer
Nov 28, 2015

Answer:

Maxima=19 at x=-1
Minimum=-89 atx=5

Explanation:

#f(x) = x^3-6x^2-15x+11#

To find the local extrema first find the critical point

#f'(x) = 3x^2-12x-15#

Set #f'(x)=0#

#3x^2-12x-15#=0

#3(x^2-4x-5)#=0

#3(x-5)(x+1)=0#

#x=5# or #x=-1# are critical points. We need to do the second derivative test

#f^('')(x)=6x-12#

#f^('')(5)=18 >0# , so #f# attains its minimum at #x=5# and the minimum value is #f(5)=-89#

#f^('')(-1) = -18 < 0# , so #f# attains its maximum at #x=-1# and the maximum value is #f(-1)=19#