= Solution
Suppose first that the unique interior minimum is nondegenerate, $\phi''(c)>0$, and $f(c)\ne0$. Outside a small neighbourhood of $c$ the phase exceeds $\phi(c)$ by a fixed positive amount, so that region is exponentially smaller. Inside, expand
$$
\phi(t)=\phi(c)+\tfrac12\phi''(c)(t-c)^2+O((t-c)^3),\qquad
f(t)=f(c)+O(t-c).
$$
The rescaling $s=\sqrt{\lambda\phi''(c)}(t-c)$ and a dominated local Gaussian calculation give the <Laplace method>
$$
\boxed{I(\lambda)\sim f(c)e^{-\lambda\phi(c)}
\sqrt{\frac{2\pi}{\lambda\phi''(c)}}.}
$$
If the minimum is at $a$ and $\phi'(a)>0$, the scale is $t-a=O(\lambda^{-1})$, giving $f(a)e^{-\lambda\phi(a)}/[\lambda\phi'(a)]$. At $b$ with $\phi'(b)<0$, replace the denominator by $\lambda|\phi'(b)|$. A stationary quadratic endpoint gives half the interior Gaussian factor.
The printed assumptions do not guarantee nondegeneracy or $f(c)\ne0$. More generally, if $\phi(t)-\phi(c)\sim A|t-c|^m$ with $A>0$ and even $m$ at an interior minimum, the leading phase scale is $\lambda^{-1/m}$ and, for $f(c)\ne0$, the factor is
$$
\frac{2f(c)\Gamma(1/m)}{m(A\lambda)^{1/m}}e^{-\lambda\phi(c)}.
$$
At a one-sided endpoint the factor two is absent and the first positive one-sided phase power need not be even. If the amplitude vanishes, expand it to its first contributing term; parity can cancel interior terms. A phase flat to all orders requires its actual local behavior rather than a nonexistent quadratic Taylor coefficient.
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