We have the cubic polynomial function
$f(x)=a x^{3}+b x^{2}+c x+d $
$f(1)=-1=a+b+c+d $
$f(2)=18=8 a+4 b+2 c+d$
$19=7 a+3 b+c(1)$
Therefore,
$f^{\prime}(x)=3 a x 2+2 b x+c $
$f^{\prime}(-1)=0 $
$3 a-2 b+c=0 $
$3 a+c=0$
$c=-3 a(2)$
and $ f^{\prime \prime}(x)=6 a x+2 b$
$f^{\prime \prime}(0)=0 $
$b=0(3)$
Substituting Eqs. (2) and (3) in Eq. (1), we get
$19=7 a -3 a$
$\Rightarrow a =\frac{19}{4}$
Therefore, $c=-\frac{57}{4}$
and $d=-1-a-b-c$
$=-1-\frac{19}{4}-0-\frac{57}{4}$
$=-\frac{(4+19+57)}{4}=-\frac{80}{4}$
Therefore, $f(x)=\frac{19}{4} x^{3}-\frac{57 x}{4}-\frac{80}{4}$
$=\frac{1}{4}\left(19 x^{3}-57 x-80\right)$
and $f ^{\prime}( x )=0 \Rightarrow \frac{1}{4}\left(57 x^{2}-57\right)=0$
$\left(x^{2}-1\right)=0 \Rightarrow x=1,-1$
$f^{\prime}(x)=\frac{57}{4}(x-1)(x+1)$
Therefore, $f ^{\prime \prime}( x )=\left(\frac{57}{2}\right) 2 x$
$\Rightarrow f^{\prime \prime}(1)>0$
The point of minima is $x=1$. Therefore,
$f(1)=\frac{1}{4}(19-57-80)=\frac{59}{2}=\frac{-59}{2}$
Point $\left(1, \frac{-59}{2}\right)$ distance from point $(-1,2)$ is
$\sqrt{(1+1)^{2}\left(\frac{-59}{2}-2\right)^{2}}=\sqrt{4+\frac{(63)^{2}}{4}}$
$=\frac{1}{2} \sqrt{63^{2}+4^{2}}$
Hence, $f(x)$ is increasing, that is, $x \geq 1$.