How do you find all critical point and determine the min, max and inflection given #f(x)=x^3-6x^2+9x+8#?

Question

1813 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-19T12:46:21

See below

Given #f(x)#, his critical point are given by #f´(x)=0#. Lets calculate

#f´(x)=3x^2-12x+9=0#

Using quadratic formula #x=(12+-sqrt(144-108))/6=(12+-6)/6#

One critical point is #x=3# and the other is #x=1#

Analyzing the sign of #f´(x)# in intervals #(-oo,1)# #(1,3)# and #(3,+oo)# we found

#f´(x)>0# in #(-oo,1)# so f is increasing there
#f´(x)<0# in #(1,3)# so f is decreasing there
#f´(x)>0# in #(3,+oo)# so f is increasing there

Sumarizing #f(x)# has a maximum in #x=1#, has a minimum in #x=3# graph{x^3-6x^2+9x+8 [-15.25, 25.33, -3, 17.27]}

If we calculate #f´´(x)=6x-12=0# we found x=2 as infexion point