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For the quartic potential,Integrating the result of part (c) along the trajectory from ${\varphi }_i$ to ${\varphi }_f$ givesBecause the negative field rolls toward zero, ${\varphi }_i<{\varphi }_f<0$ and $|{\varphi }_i|>|{\varphi }_f|$, so this monomial slow-roll e-fold count is positive.

From $3H\dot{\varphi }=-c^2{V}^{\prime }$ and $H^2=8\pi GV/(3c^2)$,The slow-roll approximation fails when this kinetic contribution is comparable with $V$, giving

The number of e-folds isUsing the estimate above,For $n=6$ and the roughly $60$ e-folds needed for the flatness problem,This is the monomial slow-roll e-fold count.