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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 :

Figure %: 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) = 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).

Figure %: 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) =    

Note that in Leibniz notation, this becomes an intuitive:

=