Let k and m be positive real numbers such that the function f(x)= {3 x^2+k √(x+1), 0<x<1; m x^2+k^2, x ≥ 1} is differentiable for all x>0. Then (8…

Let k and m be positive real numbers such that the function $\displaystyle f(x)=\left\{\begin{array}{cc}3 x^2+k \sqrt{x+1}, & 0<x<1 \\ m x^2+k^2, & x \geq 1\end{array}\right.$ is differentiable for all $\displaystyle x>0$. Then $\displaystyle \frac{8 f^{\prime}(8)}{f^{\prime}\left(\dfrac{1}{8}\right)}$ is equal to $\displaystyle \_\_\_\_$.