Show that $f(x) = x^3 -3x^2 -1 = 0$ (unique root) on the open interval $]3.1, 3.2[$ Given the function: $$ f(x)=x^3 -3x^2 -1 $$ *Show that $f(x)=0$ admits a unique root on the interval $]3.1,3.2[$I first thought of using this rule $f(a).f(b) < 0$ since the function is monotonic on the interval, but then I recalled that this rule only works for closed intervals.
Any help is appreciated
 A: By Descartes' Rule of signs, there is only one change in sign of coefficients of $f(x)$ and no change in sign of coefficients of $f(-x)$. 
So the equation $f(x)=0$ has exactly $1$ positive real root, no negative real root and $2$ complex conjugate roots.
Now use $f(a)\cdot f(b)<0$ in the interval $[3.1,3.2]$ since the root does not lie either at $x=3.1$ or $x=3.2$.
A: Rule of signes: $f(x)=x^3-3x^2-1$ has $1$ change signe and $f(-1)=-x^3-3x^2-1$ has $0$ change signe, hence $f(x)=0$ has at least $3-(1+0)=2$ imaginary roots.
    Consequently the only real root is in your open interval because of $f(3)\lt 0$ and $f(3.2)\gt 0$.  
A: Clearly $f(x)$ is continuous in $[3.1,3.2]$ and
$$ f(3.1)f(3.2)<0$$ 
by the Intermediate Mean Value Theorem, there is at least a $c\in[3.1,3.2]$ such that $f(c)=0$. Note
$$ f'(x)=3x(x-2)>0$$
in $[3.1,3.2]$, namely $f(x)$ is strictly increasing and therefore $c$ is a unique root in $[3.1,3.2]$.
A: The function $f(x)=x^3-3x^2-1$ is continuous on the whole real line, so you can apply the intermediate value theorem to any interval, for instance $[3.1,3.2]$; if you prove that $f(3.1)f(3.2)<0$, the theorem will tell you a zero exists in the open interval $(3.1,3.2)$.
Rewrite $f(x)$ as
$$
f(x)=x^3-3x^2+3x-1-3x=(x-1)^3-3x
$$
so
$$
f(3.1)=2.1^3-9.3=9.261-9.3<0,
\qquad
f(3.2)=2.2^3-9.6=10.648-9.6>0
$$
so the existence of a zero is proved.
Now $f'(x)=3x^-6x=3x(x-2)$, which is positive on $[3.1,3.2]$.
