How to find critical points and the using the first derivative test for #f(x) = 3x^4 - 4x^3 + 6#? Calculus Graphing with the First Derivative Identifying Stationary Points (Critical Points) for a Function 1 Answer Nghi N. · Kevin B. Apr 16, 2015 #f(x) = 3x^4 - 4x^3 + 6# Derivative: #f'(x) = 12x^3 - 12x^2 = 12x^2*(x - 1)# #f'(x) = 0# when #x = 0# and #x = 1#. The critical points are: #(0, 6)# and #(1, 5).# Answer link Related questions How do you find the stationary points of a curve? How do you find the stationary points of a function? How many stationary points can a cubic function have? How do you find the stationary points of the function #y=x^2+6x+1#? How do you find the stationary points of the function #y=cos(x)#? How do I find all the critical points of #f(x)=(x-1)^2#? Let #h(x) = e^(-x) + kx#, where #k# is any constant. For what value(s) of #k# does #h# have... How do you find the critical points for #f(x)=8x^3+2x^2-5x+3#? How do you find values of k for which there are no critical points if #h(x)=e^(-x)+kx# where k... How do you determine critical points for any polynomial? See all questions in Identifying Stationary Points (Critical Points) for a Function Impact of this question 1999 views around the world You can reuse this answer Creative Commons License