= Solution
The <eddy viscosity> in the prescribed <mixing length> model is
$$
\boxed{\nu_T=l^2|U_y|=C_1^2M_0^{1/2}x^{1/2}\eta^2|f''(\eta)|.}
$$
Thus at a fixed similarity coordinate $\eta$, every nonzero value grows as $x^{1/2}$. Its maximum occurs at a fixed coordinate maximizing $\eta^2|f''|$, so the maximum itself has the same downstream scaling.
The formal argument intended by the edge assumption is also clear. If one independently stipulates $|f''(\eta)|\leq|f''(\eta_w)|$ for $|\eta|\leq\eta_w$, then $\eta^2|f''(\eta)|\leq\eta_w^2|f''(\eta_w)|$. This would put the maximum <eddy viscosity> at the edge and give
$$
\frac{U(x,0)y_w}{\nu_T(x,y_w)}=\frac{f'(0)}{C_1^2\eta_w|f''(\eta_w)|},
$$
independent of $x$, provided $f''(\eta_w)\ne0$. That is a conditional scaling calculation, not an admissible solution of the full momentum problem.
The printed edge claim fails for the reason established in part (e): $f'(\eta_w)=0$ forces $f''(\eta_w)=0$, and therefore
$$
\boxed{\nu_T(x,\pm y_w)=0.}
$$
A nontrivial profile has positive <eddy viscosity> in the interior; hence its maximum cannot occur at that finite free edge. Moreover $U(x,0)y_w/\nu_T(x,\pm y_w)$ has a zero denominator, so it is \b[not a finite effective Reynolds number] under the literal assumptions.
The intended downstream cancellation is valid if one uses a consistent interior reference point $\eta_*\in(0,\eta_w)$ with $f''(\eta_*)\ne0$, for example the maximum of $\nu_T$. Since $U(x,0)=M_0^{1/2}x^{-1/2}f'(0)$ and $y_w=\eta_wx$,
$$
\boxed{\frac{U(x,0)y_w}{\nu_T(x,\eta_*x)}
=\frac{f'(0)\eta_w}{C_1^2\eta_*^2|f''(\eta_*)|},}
$$
which is independent of $x$. This is a qualified correction of the reference point, not a proof of the false edge maximum.
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