Let $f(x)$ be polynomial of degree 3 such that $f(3)=1, f^{\prime}(3)=-1, f^{\prime \prime}(3)=0$ and $f^{\prime \prime \prime}(3)=12$. Then, the value of $f^{\prime}(1)$ is

23

Let $f(x)=a(x-3)^3+b(x-3)^2+c(x-3)+d$ be the given polynomial. Then,

$f(3)=1 ~~\Rightarrow d=1$

$f^{\prime}(3)=-1 ~~\Rightarrow c=-1$

$f^{\prime \prime}(3)=0 ~~\Rightarrow b=0$

and, $f^{\prime \prime \prime}(3)=12 ~~\Rightarrow 6 a=12 \Rightarrow a=2$

∴  $f(x)=2(x-3)^3-(x-3) +1$

$\Rightarrow f^{\prime}(x)=6(x-3)^2-1 ~~\Rightarrow f^{\prime}(1)=23$