It is given that for the function

The given function is $f(x)=x^{3}-6 x^{2}+a x+b$.

It is given that Rolle's theorem holds for $f(x)$ defined on $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}$.

$f(1)=f(3)=0 \quad$ (Given)

$\therefore f(1)=0$

$\Rightarrow 1-6+a+b=0$

$\Rightarrow a+b=5$           ......(1)

Also,

$\Rightarrow 27-54+3 a+b=0$

$\Rightarrow 3 a+b=27$             ......(2)

Solving (1) and (2), we get

$a=11$ and $b=-6$

It can be verified that for $a=11$ and $b=-6, f^{\prime}\left(2+\frac{1}{\sqrt{3}}\right)=0$.

Thus, the values of a and b are 11 and −6, respectively.

It is given that for the function $f(x)=x^{3}-6 x^{2}+a x+b$ on $[1,3]$, Rolle's theorem holds with $c=2+\frac{1}{\sqrt{3}} .$ If $f(1)=f(3)=0$, then a = ___11___, b =___−6___.