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

Question

Emily Ammerman · Accepted Answer

#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.

Samantha Adkison · Answer

undefined To find the critical points of $ f(x) = x^4 $, we first find the derivative:

$$ f'(x) = 4x^3 $$

Critical points occur where the derivative is equal to zero or undefined. Here, the derivative is never undefined, so we set it equal to zero:

$$ 4x^3 = 0 $$

This gives us $ x = 0 $ as the only critical point.

To determine whether $ x = 0 $ corresponds to a minimum, maximum, or inflection point, we use the second derivative test or examine the behavior of the function around $ x = 0 $. Since the second derivative $ f''(x) = 12x^2 $ is positive for $ x = 0 $, $ x = 0 $ corresponds to a local minimum.

There are no inflection points because the function $ f(x) = x^4 $ is a polynomial of degree 4, and it does not change concavity.