How do you find the maximum value of 3x^2-x^3? Calculus Graphing with the First Derivative Identifying Stationary Points (Critical Points) for a Function 1 Answer Eddie Aug 11, 2016 4 Explanation: y = 3x^2 - x^3 y' = 6x - 3x^2 = 3x(2 - x) y' = 0 implies x = 0, 2 y'' = 6 - 6x = 6(1-x) y''(0) = 6 > 0 a min y''(2) = -6 < 0 ie a max y(2) = 4, the max value of y graph{3x^2 - x^3 [-12.66, 12.66, -6.33, 6.33]} 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 7168 views around the world You can reuse this answer Creative Commons License