Find the points that have the common tangent line

There are two points on the curve y = x^4 - 2x^2 - x that have a common tangent line. Find those points.

So far I think I know what needs to be found but I cannot put the pieces together.

y-prime = 4x^3 - 4x - 1

i need a y prime that yields a slope such that the equation y=mx+b will intersect the initial curve such that the new x value will give the same derivative value as the first point (thus the same sloped tangent)

but i do not know how to do this

Re: Find the points that have the common tangent line

Sorry for bumping such an old thread – I was solving a similar problem on another forum when I actually found this thread on Google.

Here’s how I do it.

Let $\displaystyle f(x)=x^4-2x^2-x$. If the equation of the common tangent is $\displaystyle y=mx+c$ we have

$\displaystyle m=f'(x)=4x^3-4x-1$

$\displaystyle c=f(x)-mx=x^4-2x^2-x-(4x^3-4x-1)x=-3x^4+2x^2$

We want to find two distnct points at $\displaystyle x=a$ and $\displaystyle x=b$ such that

$\displaystyle 4a^3-4a-1=4b^3-4b-1\ \Rightarrow\ a^3-b^3=a-b\ \Rightarrow\ a^2+ab+b^2=1\ \ldots\ \fbox1$

and

$\displaystyle -3a^4+2a^2=-3b^4+2b^2\ \Rightarrow\ 3(a^4-b^4)=2(a^2+b^2)$ $\displaystyle \Rightarrow\ 3(a+b)(a^2+b^2)=2(a+b)$

We will show that $\displaystyle a+b=0$. If $\displaystyle a+b\ne0$ then $\displaystyle 3(a^2+b^2)=2\ \Rightarrow\ a^2+b^2=\frac23$; substituting into $\displaystyle \fbox1$ gives $\displaystyle ab=\frac13$; also $\displaystyle (a+b)^2=a^2+b^2+2ab=\frac43$ and so $\displaystyle a+b=\frac{\pm2\sqrt3}3$. Hence $\displaystyle a,b$ are roots of the quadratic equation $\displaystyle 0=3x^2\pm2\sqrt3x+1=\left(\sqrt3x\pm1\right)^2$, i.e. $\displaystyle a=b$. We don’t want this as $\displaystyle a,b$ must be distinct.

Therefore $\displaystyle a+b=0$. Substituting $\displaystyle b=-a$ into $\displaystyle \fbox1$ gives $\displaystyle a^2=1$, i.e. $\displaystyle a=\pm1$.