Calculus Of One Real
Variable – By Pheng Kim Ving 
3.4 
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1. OneToOne
Functions 

Fig. 1.1 f is onetoone. 

Fig. 1.2
g isn't onetoone. 
function, one x corresponds to
exactly one y. Now we also have that one y corresponds to
exactly one x. Thus, f is said
to be onetoone. A horizontal line cuts the graph of f
at at most one point.
Note that the implication:
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2. Inverse Functions 
Let y = f(x)
be a function. We see that f maps x to y.
For each y in range ( f ) there correspond one or more x
in
dom( f )
where y = f(x). This correspondence maps y
back to x. So its action is the reverse of that of f.
Thus it's called
the inverse relation of f.
ie, x is the image of y by f ^{–1} iff y is the image of x by f, or f ^{–1} maps y to x iff f maps x to y.
We get a new function, f ^{–1}. Usually we treat functions as
mapping x to y. Now, f ^{–1} is a function
in its own right. Hence
let's take it out of the original situation where it's treated as mapping y
to x and put it with all other functions and treat it
as mapping x to y. It follows that we usually write y
= f ^{–1}(x),
rather than x = f ^{–1}( y), to show the mapping action of f ^{–1}.
Therefore, given that f is a onetoone function, the definition
of the inverse function f ^{–1} of f
is usually presented as:

ie, y is the image of x by f ^{–1} iff x is the image of y by f, or f ^{–1} maps x to y iff f maps y to x.
We say that a function is invertible if it has its inverse
function, ie, if its inverse relation is also a function. We've seen
that if a function is onetoone, then it's invertible. If a function y
= f(x) isn't onetoone, then there exist
two different x_{1}
and x_{2}
such that f(x_{1}) = f(x_{2}) = y_{1}. Clearly its
inverse relation isn't a function, because y_{1} would have two different
images: x_{1}
and x_{2}.
So if a function isn't onetoone then it isn't invertible. Thus a function is
invertible iff it's onetoone.
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3. Finding Inverse Functions 
Example 3.1
Let f(x)
= 2x + 1.
a. Show that f
is invertible.
b. Find its inverse, f ^{–}^{1}.
Solution
a. If f(x_{1}) = f(x_{2}), then 2x_{1} + 1 = 2x_{2} + 1, yielding x_{1} = x_{2}. So f
is onetoone, thus invertible.
b. Let y = f ^{–1}(x). Then x = f( y) = 2y + 1, yielding y = (x – 1)/2. As a consequence:
EOS
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4. Graphs Of Inverse Functions 
As shown in Example
3.1, the inverse of f(x) = 2x + 1 is f^{
–}^{1}(x)
= (x – 1)/2. The graphs of f and f ^{–}^{1} are sketched
in Fig.
4.1. They're mirror images of each other in the line y = x.
We now show that the graph of the
inverse f ^{–}^{1} of any invertible function f is the mirror image
of that of f in the line y =
x.
See Fig. 4.2. The graph of y = f(x) is the same as that of x
= f ^{–}^{1}( y) . Since the function x
= f ^{–}^{1}( y) maps y to x,
its

Fig. 4.1
Graphs of y = 2x + 1 and y
= (x – 1)/2 are mirror images of 

Fig. 4.2
Graphs of y = f(x)
and y = f ^{–1}(x)
are mirror images of each other 
domain is on the yaxis
and its range on the xaxis. Move its domain to the xaxis
and its range to the yaxis, and we get
the domain on the xaxis and the range on the yaxis of the
function y = f ^{–}^{1}(x).
This movement is done by flipping the
plane of the axes over about the line y = x, taking the domain, range, and graph
of x = f ^{–}^{1}( y)
along but leaving the
axes intact. The graph of x = f ^{–}^{1}( y)
then becomes that of y = f ^{–}^{1}(x).
Thus, the graph of y = f ^{–}^{1}(x)
is the mirror image
of that of x = f ^{–}^{1}( y),
hence of that of y = f(x), in the line y = x.
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5. Differentiation Of Inverse Functions 
Consider the inverse f ^{–}^{1} of an
invertible function f. The graph of y = f ^{–}^{1}(x)
is sketched in Fig. 5.1. The graph of x = f( y)
is the same as that of y = f ^{–}^{1}(x).
If y changes twice as fast as x does, then x
changes half as fast as y does. The rate of

Fig. 5.1 Rate of change of y with respect to x
is reciprocal of that of 
change of y with respect to x
is the reciprocal of the rate of change of x with respect to y. Now, y
is the function of x by
f ^{–}^{1}, and x is
the function of y by f. So the rate of change of y
with respect to x is ( f ^{–}^{1})'(x)
and the rate of change of x
with respect to y is f '( y). Thus, ( f ^{–}^{1})'(x)
= 1/f
'( y) = 1/f
'( f ^{–}^{1}(x)).

Using the Leibniz notation for ( f ^{–}^{1})'(x) = 1/f '( f ^{–}^{1}(x)) we obtain dy/dx = 1/f '( y) = 1/(dx/dy):

The notation dy/dx,
again, appears to be a normal fraction. This formula states that the rate of
change of y with respect
to x is the reciprocal of the rate of change of x
with respect to y.
Solution
Problems & Solutions 
1. Suppose f is an invertible function with f(1)
= – 3 and f(5) = 6. Find:
a. f ^{–1}(–3).
b. f( f ^{–1}(6)).
c. f ^{–1}(
f(1)).
a. Since f(1) = – 3 we have f ^{–1}(– 3) =
1.
b. Since f(5) = 6 we have f ^{–1}(6) = 5.
So f( f ^{–1}(6)) = f(5) = 6.
c. f ^{–1}( f(1)) = f ^{–1}(– 3) =
1.
2. Let f(x) = x^{3} + 1. Find the
derivative of the inverse of f at x = 9 in two ways:
a. By determining the inverse function and
differentiating it.
b. By using the formula for the derivative of
an inverse function.
Solution
a. Let y = f ^{–1}(x). Then x = f( y) = y^{3} + 1. So f ^{–1}(x) = y = (x – 1)^{1/3}. Thus:
Remark
However, this alternative may be
confusing, because the question uses x, not y, as the independent
variable for f ^{–}^{1} (it
says “ at x
= 9 ”, not “ at y
= 9 ” ).
3. Let f(x) = x^{n} where x > 0 is a positive real number and n
> 0 is a positive integer.
a. Show that f ^{–1}(x) = x^{1/}^{n}.
b. Differentiate f^{ –1} by using the power rule and using
the Leibniz notation.
c. Differentiate f ^{–1} by using the formula for the
derivative of an inverse function and using the Leibniz notation.
Solution
a. Let y = f ^{–1}(x). Then x = f( y) = y^{n}, so that f ^{–1}(x) = y = x^{1/}^{n}.
4. Let g(x) = (x – 1)/(x
+ 2). Calculate the derivative of the inverse of g at x
= 0 in two ways:
a. By determining the inverse function and
differentiating it.
b. By using the formula for the derivative of
an inverse function.
Solution
5. Suppose f is a onetoone differentiable function satisfying f '(x) = 1/x. Let y = f ^{–}^{1}(x). Prove that dy/dx = y.
Solution
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