How do you sketch the graph #y=x^4-27x# using the first and second derivatives?

1 Answer
May 25, 2017

Please see below.

Explanation:

As #y=x^4-27x#

observe that #x^4-27x=x(x^3-27)# and as #x^4# is always positive and rises faster than #27x#, curve, it is negative only between #0 < x < 3# and elsewhere it is positive.

#(dy)/(dx)=4x^3-27# and #(d^2y)/(dx^2)=12x^2#

Now #(dy)/(dx)=0# when #4x^3-27=0#

i.e. #(root(3)4x)^3-3^3=0#

or #(root(3)4x-3)((root(3)4x)^2+3root(3)4x+9)=0#

or #(root(3)4x-3)(root(3)16x^2+3root(3)4x+9)=0#

And #(dy)/(dx)=0# when #root(3)4x-3=0# or #x=3/root(3)4=1.8899#

and as at this point #(d^2y)/(dx^2)>0#, we have a minima.

And for #root(3)16x^2+3root(3)4x+9)=0#, discriminant is #9root(3)16-4xx9root(3)16=--27root(3)16 <0# and no other solution for #(dy)/(dx)=0# exists.

graph{x^4-27x [-10, 10, -60.2, 99.8]}