What are the roots of this function with absolute values?

Question

The function is the following $$f(x)= (\lvert x\rvert + x^2)e^{-x}$$ I don't understand how to deal with the $\lvert x\rvert$

Answer 1

Note that $f(x) >0$ for $x \neq 0$ and $f(0) = 0$.

Answer 2

One way to find the root is by reasoning. You know that $e^{-x}>0$ for all values of $x$, so that means that for your function $f(x)$ to have a root, the other factor $|x|+x^2$ must equal zero at a point. By inspection, we can see there is one value that satisfies this, and since $|x|+x^2$ will never be negative and is not periodic, there is only one root.

Answer 3

You have $$f(x)= (\lvert x\rvert + x^2)e^{-x} = \lvert x\rvert\, (\lvert x\rvert + 1)\,e^{-x}.$$

Since $ (\lvert x\rvert + 1)\,e^{-x} > 0$ for $x \in \mathbb{R}$, the only way to get $f(x)=0$ is to have

$$\lvert x\rvert = 0.$$

This is by definition of absolute value only when $x=0$ (see here).

Answer 4

We note that $$f(x)=(|x|+x^2)e^{-x}=\begin{cases}(x+x^2)e^{-x} & ,x > 0 \\ (-x+x^2)e^{-x} & ,x \le 0\end{cases}$$

$e^{-x}>0 \, \forall \, x \ge 0$ and hence $f(x) > 0 \, \forall \, x \ge 0$.

At $x=0$, $f(x)=0$.

The only solution hence is $x=0$.

Hope this helps you.

Answer 5

If $x$ is a complex number, there are two more solutions.
Firstly, $e^{-x}$ is never zero because $|e^{-x}|=e^{-Re(x)}>0$.
So the only way to get zero is if $|x|+x^2=0$.
Let $x=re^{i\theta}$, then $|x|+x^2=r+r^2e^{2i\theta}$. Then either $r=0$, so $x=0$ which you already have; or $re^{2i\theta}=-1$. Then $r=1,\theta=\pm\frac{\pi}2$, and $x=\pm i$