How do you find all critical point and determine the min, max and inflection given #f(x)=x^4#?

1 Answer
Oct 26, 2016

#f(x)=x^4#

Domain of #f# is #(-oo,oo)#.

A critical number for #f# is a number #c# in the domain of #f# at which #f'(c)# does not exist or #f'(c)=0#

#f'(x) = 4x^3# is defined for all #x# and is #0# at #x=0#.

The only critical number for #f# is #0#.

(If your treatment of calculus says that a critical point is a point on the graph, then the critical point is #(0,0)#)

Because #f'(x)# is negative for #x < 0# and positive for #x > 0#, we know that #f(0) = 0# is a local minimum.

#f''(x) = 12x^2# which is always positive.

Since the sign of #f''# never changes, the concavity of #f# never changes, so there are no inflection points.