So I had these two functions and the following exercise:
$f(x)=x^3-2x^2$
$g_p(x)=px$
Determine all values of $p$ for which $f$ and $g_p$ have three points in common.
Whats a good way to tackle this? I tried equaling them to each other but that doesn't seem to work. Any good methods? (No hints please).
On
Yes, equalling them is the way to go. Then use the fact that
A polynomial $P$ of degree $3$ has three distinct roots if and only if its derivative $P'$ vanishes in two points $x_1,x_2$ such that $P(x_1)$ and $P(x_2)$ have opposite sign.
The derivative of $P$ here is $3x^2-2x-p$. The quadratic formula tells you what the roots are. Finally, note that $P(x_1)$ and $P(x_2)$ have opposite sign is equivalent to $$P(x_1)P(x_2)<0$$ Therefore all you have to do is solve this inequation for $p$.
Edit: @JoséCarlosSantos method is more efficient here since you can factor out an $x$. I'll leave this method though since it may apply in a more general context.
On
We have: $$x^3+2x^2=px$$ We can rearrange this to: $$x(x^2+2x-p)=0$$ From this we gather $x=0$ is always a solution.
We then need to find the $p$ such that $x^2+2x-p$ has two real solutions.
We use the Discriminant $D=b^2-4ac$ for this, since the solutions to $ax^2+bx+c=0$ are found by: $$x=\frac{-b\pm\sqrt D}{2a}$$ and $D$ must be $>0$ for the solutions to be real and distinct. Here $a=1, b=2, c=-p$, and so we get: $$2^2-4(1)(-p)>0$$ $$\to4+4p>0$$ $$\to p>-1$$
The method that you mentioned is the way to go:\begin{align}x^3-2x^2=px&\iff x=0\vee x^2-2x=p.\end{align}If $p=0$, then there will be only two points in common ($(0,0)$ and $(2,0)$). So, solve the equation $x^2-2x=p$ assuming that $p\neq0$ and make sure that you get two distinct solutions.