Let $E^\bullet$ be a perfect complex of $R$-modules (where $R$ comm. ring). So $E^\bullet$ is quasi-isomorphic to a bounded complex of finitely generated projective R-modules. Now $E^\bullet$ has Tor-amplitude in some interval $[a,b]$, which means that $H^i(E^\bullet \otimes_R M) = 0$ for all $i \not\in [a,b]$ and for all $R$-modules $M$.
Now my question is, let $(E^\bullet)^\vee$ be the dual of $E^\bullet$. What can we say about the Tor-amplitude of $(E^\bullet)^\vee$? Of course $H^i((E^\bullet)^\vee \otimes_R M) = H^i (RHom_R(E^\bullet, M)) =: Ext^i(E^\bullet, M)$. so the question is about which degrees the Ext is concentrated in. my guess is $[-b,-a]$.
Your guess is correct. I think that the easiest way to prove this is to note that for some $c,d$, $E^\bullet$ is quasi-isomorphic to a bounded complex of finitely generated projectives concentrated in degrees $[c,d]$ such that the first differential is not a split monomorphism and the last differential is not a (split) epimorphism, and that then $-d,-c$ are precisely the smallest and largest integers $i$ such that $Ext^i(E^\bullet, M)$ does not vanish for all $M$ (take $M$ to be the cokernel of the last non-zero differential or the first non-zero term of $E^\bullet$ respectively).
And dualizing turns $E^\bullet$ into a complex concentrated in degrees $[-d,-c]$ with exactly the same properties.