How do you use the first and second derivatives to sketch #y= 3x^4 - 4x^3#?

Question

Ezra Adams · Accepted Answer

See answer below

Given: #f(x) = y = 3x^4 - 4x^3 = x^3(3x - 4)# Find the first derivative and set #y' = 0# to find the critical points: # " "y' = 12x^3 - 12x^2 = 0# #" "y' = 12x^2(x - 1) = 0# #" "#critical values: #" "x = 0, x = 1# #" "f(0) = 0; " " f(1) = 3(1)^4 - 4(1)^3 = -1# #" "#critical points: #(0, 0), (1, -1)# Find the second derivative and evaluate it at the critical values to find relative minimums or relative maximums: #y'' = 36x^2-24x # #y''(0) = 0 " "=> #need to use the first derivative test to find out if this point is a relative min., max. or inflection. #y''(1) = 36(1)^2 - 24(1) = 12 > 0 => #relative min. at #(1, -1)# Find inflection points by setting #y'' = 0#: #y'' = 36x^2-24x =0# #y'' = 12x(3x - 2) = 0 =># inflections at # x = 0, x = 2/3# #f(2/3) = 3(2/3)^4 - 4(2/3)^3 = 3*16/81 - 4*8/27 = -16/27# inflection points #(0, 0), (2/3, -16/27)# To graph: Fourth order functions have end conditions both positive. Put a point at #(0,0)# and a point at the relative minimum #(1, -1)# Start at the upper left and create a horizontal inflection at #(0.0)#, then turn downward to the relative minimum, then curve upward. graph{3x^4-4x^3 [-5, 5, -5, 5]}

Adam Adkins · Answer

undefined 1. Find critical points by setting the first derivative equal to zero and solving for x.
2. Determine the nature of each critical point using the second derivative test.
3. Sketch the graph using the critical points and information about the concavity from the second derivative.