This is the potential of <Starobinsky inflation>. Put $b=\sqrt{2/3}/M_{\rm Pl}$ and $y=e^{-b\phi}$, so $V=V_0(1-y)^2$. Its <potential slow-roll parameter> and <second potential slow-roll parameter> are
$$
\epsilon_V=\frac{M_{\rm Pl}^2}{2}\left(\frac{V'}V\right)^2
=\frac43\frac{y^2}{(1-y)^2},
$$
$$
\eta_V=M_{\rm Pl}^2\frac{V''}V
=\frac43\frac{-y+2y^2}{(1-y)^2}.
$$
At large positive $\phi$, $y\ll1$, so both $\epsilon_V$ and $|\eta_V|$ are small and <slow-roll inflation> is possible. Near the minimum, a <Taylor expansion> gives $1-y\simeq b\phi$ and hence
$$
\epsilon_V\simeq\eta_V\simeq\frac{2M_{\rm Pl}^2}{\phi^2}.
$$
They are large for $|\phi|\ll M_{\rm Pl}$, so that region cannot sustain slow roll.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-310/2/c/solution