Question

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

Answer 1

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]}