Show that $c$ is in interval $[e,3]$ for $c\cdot \ln{c} + c − 6 = 0$ Show that there is a unique number $c \in \mathbb{R}$ that fulfills the equation and that this number is in the interval $[e, 3]$.
$$ c\cdot \ln{c} + c − 6 = 0$$
At first I was thinking about using linear approximation, but then I would need an $a$ and $x$. So I'm quite confused as how to start solving this problem.
 A: Note $c>0$.
Set $e^y:=c$, then
1)$ye^y +e^y -6=0$,
$e^y(1+y)-6=0$.
$f(y):=(1+y)e^y-6.$
$f(0)=-5;$ 
$f'(y)=e^y+(1+y)e^y \gt 0$ for $y \ge 0$.
Strictly increasing for $y \ge 0.$
$y=1$: $f(1)= 2e-6 <0$.
$y=\log 3$:
$ f(\log 3)=(1+\log 3)3-6>0$, 
since $\log 3 >1$.
Recalling $c=e^y:$
$c \in [e,3]$.
2) Let $y \le 0$.
A) $-1 \le y \le 0.$
$f(y)=(1+y)e^y -6= (1-|y|)e^{-|y|}-6 \le -5.$
$f(0)=-5$; $f(-1)= -6.$
B) $y \lt -1.$
$f(y)= (1+y)e^y -6 \lt -6.$
Combining:
$f(y) \lt 0$ for $y \le 0.$
$f(y)$ is strictly increasing for $y \ge 0$.
Hence $f(y)$ has exactly one zero, $y \in \mathbb{R}$.
A: Let $f(x)=x\ln(x)+x-6$.
$f$ is continuous on $[e,3].$
and
$$\forall x\in [e,3] \; \; f'(x)=\ln(x)+2>0$$
$f$ is strictly increasing on $[e,3]$.
$$f(e)=2e-6\approx -0.4$$
$$f(3)=3\ln(3)-3>0$$
by IVT, $$\exists \; \color{red}{!} \;c\in[e,3] \; : f(c)=0$$
A: Let $f(x) = x + x ln(x) - 6$
Since,
$f(e)=2e-6 <0$ 
and
$f(3)=3 ( ln(3) - 1) > 0 $
and because $f$ is continuous on $ [e,3]$ and differentiable on $(e,3)$, and also $f'(x) = ln(x) +2 > 0$ (strictly increasing), using the intermediate value theorem, there is such a $c$.
Note: The uniqueness comes from the fact that $f$ is strictly increasing.
