<s>[INST] Range of $z,$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$. Then value of $z$ lie in the interval.
My Try:: Let $x,y,z$ be the roots of the quadratic equation 
$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$
Let $xyz = p, $ Then let  $f(t) = t^3-3t^2-9t-p.$
Now we will check the extermes value of $f(t)$
$f^{'}(t) = 3t^2-6t-9 = 3(t^2-2t-3)$
$f^{''}(t) = 3(2t-3)$
Now for Max. or Min. , $f^{'}(t)=0$ or $t=-1\;,t=3$
So $f^{''}(-1) = -15=-$(ve)
so Local Max. at $t=-1$
and $f^{''}(3) = 9 = +$(ve)
So Local Min. at $t=3$
Now My question is How can I check Range of $z$ using Graph of cubic equatio using Derivatives
Thanks
 [/INST] Equating the values of $x$
$$3-y-z=-\frac{9+yz}{y+z}\implies y^2+y(z-3)+z^2-3z-9=0$$
As $y$ is real, the discriminant $(z-3)^2-4\cdot1\cdot(z^2-3z-9)$ must be $\ge0$
$$\implies z^2-2z-15\le0\implies (z-5)(z+3)\le0$$
As $(x-a)(x-b)\le 0$ for $a\le b\implies a\le x\le b$
So,$-3\le z\le 5$
Observe that the given conditions are symmetric wrt $x,y,z$
So, $x,y,z$ will have the same range.
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