<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] We have $x+y+z = 3$ and $xy+yz+zx = -9$
$xy+z(3-z)=-9$
$\Rightarrow xy+3z-z^2=-9$
$\Rightarrow xy+3z-z^2+9=0$
$\Rightarrow -xy-3z+z^2-9=0$
$\Rightarrow z^2-3z-9=xy$.....(1)
By A.M G.M inequality we have,
$\displaystyle |xy|\le(\frac{x+y}{2})^2=(\frac{3-z}{2})^2$
Using this in 1 we have,
$-(\frac{3-z}{2})^2\le z^2-3z-9\le (\frac{3-z}{2})^2$
$4(z^2-3z-9)\le (z^2-6z+9)$
$3z^2-6z-45\le 0$
$z^2-2z-15\le0$
Evaluate this and the left part and take the intersection of the two answers as your answer.
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