More formally, if f is a one-to-one function with domain D and range R, then its inverse f^-1 has domain R and range D. f^-1 is related to f in the following way: If f (x) = y, then f^-1(y) = x. Written another way, f^-1(f (x)) = x.

Example: f (x) = 3x - 4. Find f^-1(x).

The procedure for finding f^-1(x) from f (x) involves first solving for x in terms of y.

y		= 3x - 4
x		=

Now switch the variables x and y in the equation to generate the inverse:

y		=
f^-1(x)		=

A function and its inverse are related geometrically in that they are reflections about the line y = x:

Figure 1.1: A function and its inverse are symmetric with respect to the line y = x

Thus, if (a, b) is a point on the graph of f, then (b, a) is a point on the graph of f^-1.

The Derivative of the Inverse

Drawn below is the graph of f (x) = x² on the interval (0,∞), and its inverse on that interval, f^-1(x) =

. Also drawn on the graph are the tangents to the graph of f (x) at (2,4), and the tangent to the graph of f^-1(x) at the reflected point (4,2).

Figure 1.2: The tangent lines drawn at corresponding points on the graphs of f and f^-1

What is the relationship between f (x) at (a, b) and f^-1(x) at (b, a)?

In the case above, f'(x) = 2x and (f^-1)'(x) =

It seems that at least in this case, the derivative of f at (a, b) is the reciprocal of the derivative of f^-1 at (b, a). This in fact holds true in all cases. In general, it can be said that if (a, b) is a point on f then (b, a) is a point on f^-1, and (f^-1)'(b) =

To make this statement even more applicable, we should now try to find a formula for (f^-1)'(x). From the formula above, if we let b = x, then a = f^-1(x), so that the following more general statement may be written:

(f^-1)'(x) =