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

1 Answer
May 23, 2018

See answer below

Explanation:

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]}