Chapter 3, Problem 21P

To determine

To find:

The two points on the curve $y=x^4-2x^2-x$ .

Answer to Problem 21P

The two points are $\left(1,-2\right),\left(-1,0\right)$ .

Explanation of Solution

Given:

The function is

  $y=x^4-2x^2-x$

Concept used:

The slope of the tangent to a curve $f\left(x\right)$ at any point a is given by the value of the first derivative of the slope at the point $x=a$ .

The tangent to be horizontal so the slope should be equal to 0

That is $\frac{dy}{dx}=0$ .

Calculation:

The cubic function is

  $y=x^4-2x^2-x\mathrm{................................}\left(1\right)$

Differentiating equation (1) with respect to $x$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(x^4-2x^2-x\right)\\ \frac{dy}{dx}=\left(4x^3-4x-1\right)\mathrm{...................}\left(2\right)\end{array}$

From the theory of cubic $a{x}^3+b{x}^2+cx+d=0$

Here $a=4,c=0,b=-4,d=-1-m$

A double root of the equation in term of $m$

Two distinct points having the same $m$ value

  $\Delta =0$

  $\begin{array}{l}256+256m+432{\left(m+1\right)}^2=0\\ m=-1\end{array}$

The slope of $m=-1$ satisfies the equation (1)

Solving the equation

  $4x^3-4x-1$

The value of $x=0,1,-1$

The two points are $\left(1,-2\right),\left(-1,0\right)$