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Every one-to-one function f has an
inverse function f^{-1} which essentially reverses the operations performed by
f.

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:

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) = x^{2} 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).

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) =

Note that in Leibniz notation, this becomes an intuitive: