hey guys! I've got another question!
how do i find the x intercepts, by hand, if f(x)=|x|^2-4|x|
i know you make f(x)=0 and then solve but i just dont know the process.
Edit: I way over-complicated the problem. There's clearly a much better method... but I'll leave this here to be laughed at lol
\[f(x)=x^2-4|x|\] (Note is that \(|x|^2=x^2\). The absolute value brackets are redundant).
To proceed with solving equations involving the absolute value function, we generally want to break it up into cases. In some cases like this one, you can actually make a few clever observations to arrive at the answer quicker, but I'll show that process at the end.
Recall that the definition of the absolute value function is: \[|x|=\begin{cases}x, & x\geq 0\\
-x, & x<0\end{cases}.\] In our case, we would like to break up the problem into the cases \(x\geq0\) and \(x<0\).
Case 1: \(x\geq 0\)
\[x^2-4x=0\implies x(x-4)=0\implies \ \ x=0,\ 4\]
Case 2: \(x<0\) \[x^2-4(-x)=0\implies x(x+4)=0\implies \ \ x=-4\]
Putting our results together, we see that the solutions to \(f(x)=0\) are \(x=-4,\ 0,\ 4\).
Some clever observations we can make to make our lives easierNote that in this question, \(f(x)\) is an
even function (that is, \(f(-x)=f(x)\) ).
So, we must have \(f(a)=0\iff f(-a)=0\).
This means, we could've just considered
Case 1, and realise that since \(x=4\) is a solution, then so must \(x=-4\).