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If $f$ has one root on $(-\infty,0)$ and two distinct roots on $(0,+\infty)$ and $f(0)=-1$, how many roots does $|f(|x|)|$ have?

I know graph of $|f(|x|)|$ should be in quadrant I because $x$ and $y$ have positive values only but I don't know what exactly happens to graph of $f$. Should it be symmetric with respect to the y-axis? So roots on $(-\infty,0)$ won't be roots? It'd have two distinct roots, Am I right?

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It has two distinct roots in $(0,\infty)$. And it's symmetric with respect to the $y$-axis. Therefore... – TonyK Jan 24 '12 at 8:34
up vote 3 down vote accepted

For all real numbers $x$, $\: |f(|x|)| = 0 \:$ if and only if $\: f(|x|) = 0 \:$. $\:\:$ $f(|x|)$ has roots at exactly those values $x$ such that $|x|$ is a root of $f$. $\:\:$ For all $x$, $\: 0\leq |x| \:$. $\:\:$ Since $0$ is not a root of $f$, there are exactly two values $|x|$ could have while being a root of $f$, and these are both positive. $\:\:$ Since each of those two values of $|x|$ is non-zero, each of those two values of $|x|$ arises from exactly two values of $x$. $\:$ Since $|x|$ is a function of $x$, those two pairs of $x$ values do not overlap.

Therefore $|f(|x|)|$ has exactly 4 roots.

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