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Inverse functions

We prove the inverse function theorem. A continuously differentiable function which has non-zero derivative at a point has an inverse in a neighbourhood of that point (see Figure 6).

Figure 6: The Inverse Function
\includegraphics{figures.6}

A function is called monotone if it preserves or reverses order over the domain of its definition. In other words if (f (x) - f (y))(x - y) does not change sign for all x and y in the domain of definition of f. We say f is strictly monotone if in addition the above product is zero only when x = y. It is clear that a strictly monotone function is one-to-one.

Exercise 46   Let f be monotone and continuous in for x in the interval a < x < b. Show that there is a function g on the range of f so that g(f (x)) = x for all x in the interval a < x < b. (Hint: Use the intermediate value theorem and extremal value theorem).

One way to ensure that a function is monotone is to use Rolle's theorem in reverse to prove:

Exercise 47   If f is continuously differentiable and f'(x0) $ \neq$ 0 then show that f is monotonic in some interval around x0. Hence show that f has a inverse g (as in the exercise above) in some small enough interval around f (x0).

We can also compute the formal inverse

Exercise 48   If f can be expressed as

f (x) = f (x0) + f1(x - x0) + ... + fn(x - x0)n + o((x - x0)n)

with f1 $ \neq$ 0, then show that the inverse function g(y) has the following form where y0 = f (x0).

g(y) = x0 + $\displaystyle {\frac{1}{f_1}}$(y - y0) - $\displaystyle {\frac{f_2}{f_1^3}}$(y - y0)2 + ... + gn(y - y0)n + o((y - y0)n)

where gn is of the form Pn(f1,..., fn)/f1n + 1, where Pn is a polynomial function.


next up previous
Next: Implicit functions Up: New Functions from old Previous: New Functions from old
Kapil H. Paranjape 2001-01-20