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

Aug 6, 2017

See below

#### Explanation:

$y = {x}^{4} - {x}^{3} - x$

Using the power rule

$y ' = 4 {x}^{3} - 3 {x}^{2} - 1$

$y ' ' = 12 {x}^{2} - 6 x$

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

$\therefore 4 {x}^{3} - 3 {x}^{2} - 1 = 0$

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 $\frac{p}{q}$, 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 $\left(4\right)$ are $1 , 2 , 4 \left[q\right]$
and factor of the trailing constant $\left(- 1\right)$ is $1 \left[p\right]$

Hence, the possible rational roots of $y '$ are $\pm \frac{1}{1} , \pm \frac{1}{2} , \pm \frac{1}{4}$

Testing each in turn reveals $y ' \left(1\right) = 4 - 3 - 1 = 0$

Hence $x = 1$ is a rational root of $y ' \to \left(x - 1\right)$ is a factor.

To find any other real roots:

$\frac{4 {x}^{3} - 3 {x}^{2} - 1}{x - 1} = 4 {x}^{2} + x + 1$

This is a quadtatic of the form: $a {x}^{2} + b x + c$

Test for real roots: ${b}^{2} - 4 a c \ge 0$

${1}^{2} - 4 \times 4 \times 1 < 0 \to$ the other two roots $\in \mathbb{C}$

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

To test the nature of $y \left(1\right)$:

$y ' ' \left(1\right) = 12 - 6 > 0 \to y \left(1\right) = {y}_{\min} = - 1$

$\therefore y$ 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, $\left(- 1.2 , + 2\right)$ to produce this graph.]