A common tangent line

Question

The graph of $f(x)=x^4+4x^3-16x^2+6x-5$ has a common tangent line at $x=p$ and $x=q$. Compute the product $pq$.

So what I did is I took the derivative and found out that $p^2+3p+q^2+3q+pq=0$. However when I tried to factorize it I didn't find out an obvious solution. Can someone hint me what to do next? Thanks in advance

Answer 1

Since the specific gradient is not given, I think it would be appropriate to leave it in a very general form. If $$ p^{2} + 3p + q^{2} + 3q + pq = 0 $$ were correct (which we have verified is not true), then $$ pq = -\left(p^{2} + q^{2} + 3p + 3q\right). $$

Out of personal preference, I would rewrite $$ p^{2} + pq + q^{2} + 3p + 3q = (p + q)^{2} - pq + 3(p + q) = 0 $$ so that $$ pq = (p + q)^{2} + 3(p + q). $$

Answer 2

Let the equation of the common tangent be $y=mx+b$.

Solving simultaneously the equations of the common tangent and $f(x)$, we get

$x^4+4x^3-16x^2+(6-m)x-(b+5)=0$

This equation will have two double roots (since a line is touching a curve twice), so let the roots be $p,\,p,\,q,\,q$.

Can you take it from here?

Answer 3

Consider the quartic polynomial $P(x)=f(x)-ax$ and let $t:=x+b$. Find $a$, $b$ such that $t\to P(t-b)$ is an even function by requiring that the coefficients of $t$ and $t^3$ are zero. It turns out that $b=1$ and $a=46$.

Hence $P(t-1)=t^4-22t^2+16$ whose derivative is $P'(t-1)=4t(t^2-11)$. Therefore $P$ has two minimum points at $t_1=-\sqrt{11}$ and $t_2=\sqrt{11}$ with a common horizontal tangent. Assuming $p<q$, it follows that $$p=t_1-b=-\sqrt{11}-1\quad\mbox{ and }\quad q=t_2-b=\sqrt{11}-1\implies pq=-10.$$

Answer 4

Hint:

If the line $y=mx+n$ is tangent of the curve $y=f(x)=x^4+4x^3-16x^2+6x-5$ at the points $P=(p,f(p))$ and $Q=(q,f(q))$, this means that the system $$ \begin{cases} y=x^4+4x^3-16x^2+6x-5\\ y=mx+n \end{cases} $$ has the two double solutions $x=p$ and $x=q$. And this means that the polynomial $x^4+4x^3-16x^2+6x-5-mx-n$ factorize as $(x-p)^2(x-q)^2$.

Using the identity principle for polynomial, from $$ x^4+4x^3-16x^2+6x-5-mx-n=(x-p)^2(x-q)^2 $$ you can find $p$ and $q$.

Note: this is a solution that does not use the derivatives (only algebra, no calculus), but you can solve the problem also noting that $y=f(x)$ has a''bi-tangent''at $x=p$ and $x=q$ iff: $$ \frac{f(p)-f(q)}{p-q}=f'(p)=f'(q)=m $$ where $m$ is the slope of the bi-tangent line.

Answer 5

The tangent line equation at $x=p$ and $x=q$ is: $$y=p^4+4p^3-16p^2+6p-5+(4p^3+12p^2-32p+6)(x-p);\\ y=q^4+4q^3-16q^2+6q-5+(4q^3+12q^2-32q+6)(x-q).\\$$ By equating the coefficients, simplifying and denoting $a=p+q, b=pq$ we get: $$\begin{cases}b=a^2+3a-8\\ 3a^3-(6a+8)b+8a^2-16a=0\end{cases}$$ Solving the system we get: $$a=-2,b=-10=pq.$$