How do you sketch the graph `y=x^4-x^3-x` using the first and second derivatives?

`y = x^4-x^3-x`

Using the power rule

`y' = 4x^3-3x^2-1`

`y'' = 12x^2-6x`

For turning points of `y`: `y'=0`

`:. 4x^3-3x^2-1 =0`

This is a polynomial of degree 3. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction `p/q`, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

In this case the factors of the leading coefficient `(4)` are `1,2,4 [q]`
and factor of the trailing constant `(-1)` is `1 [p]`

Hence, the possible rational roots of `y'` are `+-1/1, +-1/2, +-1/4`

Testing each in turn reveals `y'(1) = 4-3-1=0`

Hence `x=1` is a rational root of `y' -> (x-1)` is a factor.

To find any other real roots:

`(4x^3-3x^2-1 )/(x-1) = 4x^2+x+1`

This is a quadtatic of the form: `ax^2+bx+c`

Test for real roots: `b^2-4ac>=0`

`1^2-4xx4xx1<0 ->` the other two roots `in CC`

Hence, `x=1` is the only real root of `y'`

To test the nature of `y(1)`:

`y''(1) = 12-6 > 0 -> y(1) = y_min =-1`

`:. y` has a single minimum value of `-1` at `x=1`

By inspection it is clear that `y` has a zero at `x=0`

To find other zeros:

`x^3-x^2-1=0`

Unfortunately, this cubic has no rational roots and one real root at `x approx 1.46557` This result was obtained using an online polynomial root calculator: http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php

We now have the critical points of `y` shown on the graph below

graph{x^4-x^3-x [-10, 10, -5, 5]}

[NB: In practice, it would probably be necessary to plot a few extra points in the interval, say, `(-1.2,+2)` to produce this graph.]