Let the sum of the maximum and the minimum values of the function f(x)=(2 x^2-3 x+8)/(2 x^2+3 x+8) be m/n, where gcd ( m, n )=1. Then m + n is equal…

Let the sum of the maximum and the minimum values of the function $\displaystyle f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}$ be $\displaystyle \frac{\mathrm{m}}{\mathrm{n}}$, where $\displaystyle \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$. Then $\displaystyle \mathrm{m}+\mathrm{n}$ is equal to :