Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 6

In attempting to prove in Problem 6 that
\[\Large{\theta=\frac{\psi\left(\frac{\phi(a+h)-\phi(a)}{h}\right)-a}{h}}\]
De Morgan made the assumption that the function \(\phi'\) is always invertible in the domain under consideration. And as any good calculus student will know, this is by no means necessarily true.

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