Inverse, Exponential, and Logarithmic Functions
Inverse Functions
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:
 = 
|
