How do you find the stationary points for #f(x)=x^4# ?

Question

Evelyn Agard · Accepted Answer

The stationary point is #(0,0)# and it is a minimum point. The first task in finding stationary points is to find the critical points, that is, where #f'(x)=0# or #f'(x)# DNE: #f'(x)=4x^3# using the power rule
#4x^3=0#
#x=0# is the only solution There are 2 ways to test for a stationary point, the First Derivative Test and the Second Derivative Test. The First Derivative Test checks for a sign change in the first derivative:  on the left the derivative is negative and on the right the derivative is positive, so this critical point is a minimum. The Second Derivative Test checks for the sign of the second derivative: #f''(x)=12x^2#
#f''(0)=0# There is no sign, so the second derivative doesn't tell us anything in this case. Since there are no other minimums and #lim_(x->-oo)f(x)=lim_(x->oo)f(x)=oo#. The stationary point is an absolute minimum.

Christopher Allers · Answer

undefined To find the stationary points for $ f(x) = x^4 $, you need to find where the derivative of the function $ f'(x) $ equals zero. Taking the derivative of $ f(x) $, you get $ f'(x) = 4x^3 $. Setting $ f'(x) $ equal to zero and solving for $ x $, you find the stationary points. $ f'(x) = 0 $ implies $ 4x^3 = 0 $, which has a single solution at $ x = 0 $. Therefore, the stationary point for $ f(x) = x^4 $ is at $ x = 0 $.