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.]