What are the critical points of #f(x) = x^3 − 12x + 7#?

1 Answer
Jan 7, 2016

Answer:

The critical points are:

MIN: #(2;-9)#
MAX: #(-2;23)#

Explanation:

The critical points are the x values where:

#f'(x)=0#

We obtain:

#f'(x)=3x^(3-1)-12x^(1-1)+0=3x^2-12=3*(x^2-4)=#
#=3*(x+2)(x-2)#

Then:

#f'(x)=0 iff 3*(x+2)(x-2)=0#

a multipication is zero if the factors are zero:

#x+2=0#

#x_1=-2#

#x-2=0#

#x_2=+2#

Now, to find if they are min or max you have to evaluate #f'(x)>0#

#f'(x)>0# for #x in ]-oo;-2[ uu ]2;+oo[#

then

#f(x)# is growing for #x in ]-oo;-2[ uu ]2;+oo[#

#f'(x)<0# for #x in ]-2;2[ #

then

#f(x)# is decreasing for #x in ]-2;2[#

then
for #x=-2# we have a MAX
for #x=2# we have a MIN

#f(-2)=-8+24+7=23#

#f(2)=8-24+7=-9#