smc

Since $x^3=-(a_2{x}^2+a_{1}x+a_0)$ and $x$ will be as big as possible, we need $x^3$ to be as big as possible, which means $a_2{x}^2+a_{1}x+a_0$ is as small as possible. Since $x$ is positive (according to the options), it makes sense for all of the coefficients to be $-2$. Evaluating $f(\frac{5}{2})$ gives a negative number, $f(3)$ 1, and $f(\frac{7}{2})$ a number greater than 1, so the answer is $\boxed{\frac{5}{2}<r<3}$