What are the critical points of #f(x) = -3x^4 + 12x^3 + 6#? Calculus Graphing with the First Derivative Identifying Stationary Points (Critical Points) for a Function 1 Answer Eddie Jul 4, 2016 #x = 0# an inflexion #x=3# a max Explanation: #f(x) = -3x^4 + 12x^3 + 6# #f'(x) = -12x^3 + 36x^2 = 12x^2(3-x)# #f'' (x)= -36x^2 +72x = 36x(2-x) # #f' = 0 \implies x = 0, 3# #f''(0) = 0# ie an inflexion #f''(3) = -108# ie a max 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 2440 views around the world You can reuse this answer Creative Commons License