We have,
$\begin{aligned} & f(x)=\sqrt{x^2-x}, x \in[1,4] \Rightarrow f(1)=\sqrt{1-1}=0 \\ & f(4)=\sqrt{16-4}=\sqrt{12}=2 \sqrt{3} \Rightarrow f^{\prime}(c)=\frac{2 c-1}{2 \sqrt{c^2-c}}\end{aligned}$
By Lagrange's mean value theorem
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
$\frac{2 c-1}{2 \sqrt{c^2-c}}=\frac{f(4)-f(1)}{4-1}=\frac{2 \sqrt{3}-0}{3}$
$\Rightarrow \frac{2 c-1}{2 \sqrt{c^2-c}}=\frac{2}{\sqrt{3}}$
$\Rightarrow \quad \sqrt{3}(2 c-1)=4 \sqrt{c^2-c} \Rightarrow 3(2 c-1)^2=16 c^2-16 c$
$\begin{aligned} & \Rightarrow 3\left(4 c^2-4 c+1\right)=16 c^2-16 c \\ & \Rightarrow 12 c^2-12 c+3=16 c^2-16 c \Rightarrow 4 c^2-4 c-3=0 \\ & \Rightarrow(2 c-3)(2 c+1)=0 \Rightarrow c=3 / 2 \in(1,4)\end{aligned}$
$\therefore \quad c=3 / 2$