\documentclass{amsart}
\begin{document}
Define $\sigma_2, \sigma_3$ by: $w-\mu_1 = \sigma_2$, $\mu_4-w = \sigma_3$.
\\Set $f'_2(w) = f'_3(w)$: $-\frac{a_2(w-\mu_2)}{\sigma_2^2}e^{-\frac{1}{2}} = -\frac{a_3(w-\mu_3)}{\sigma_3^2}e^{-\frac{1}{2}}$
\\$a_2/\sigma_2 = -a_3/\sigma_3$
\\Setting $f_2(w) = f_3(w)$ gives $b_2+a_2e^{-1/2} = b_3+a_3e^{-1/2}$. Thus $b_3 = b_2+a_2e^{-1/2}-a_3e^{-1/2} = b_2+a_2e^{-1/2}+a_2e^{-1/2}\sigma_3/\sigma_2$
\\At $\mu_1$, both $f_1'$ and $f_2'$ are $0$, so we only need to match $f_1$ and $f_2$.
\\Get that $b_1+a_1 = b_2+a_2$.
\\Similarly, $b_3+a_3 = b_4+a_4 = b_2+a_2e^{-1/2}(1+\sigma_3/\sigma_2)-a_2\sigma_3/\sigma_2 = b_1+a_1+a_2(1+\sigma_3/\sigma_2)(e^{-1/2}-1)$.
\\So $a_2 = \frac{b_4+a_4-b_1-a_1}{(1+\sigma_3/\sigma_2)(e^{-1/2}-1)}$
\\Similarly, $a_3 = \frac{b_1+a_1-b_4-a_4}{(1+\sigma_2/\sigma_3)(e^{-1/2}-1)}$
\\So we get
\\$f_1 = b_1+a_1e^{-\frac{1}{2}\left(\frac{x-\mu_1}{\sigma_1}\right)^2}$
\\$f_2 = b_1+a_1+(w-\mu_1)\frac{b_4+a_4-b_1-a_1}{(\mu_4-\mu_1)(e^{-1/2}-1)}(e^{-\frac{1}{2}\left(\frac{x-\mu_1}{w-\mu_1}\right)^2}-1)$
\\$f_3 = b_4+a_4+(\mu_4-w)\frac{b_1+a_1-b_4-a_4}{(\mu_4-\mu_1)(e^{-1/2}-1)}(e^{-\frac{1}{2}\left(\frac{x-\mu_4}{w-\mu_4}\right)^2}-1)$
\\$f_4 = b_4+a_4e^{-\frac{1}{2}\left(\frac{x-\mu_4}{\sigma_4}\right)^2}$
\end{document}