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Calculus Of One Real
Variable – By Pheng Kim Ving |
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3.4 |
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1. One-To-One
Functions |
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Fig. 1.1 f is one-to-one. |
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Fig. 1.2
g isn't one-to-one. |
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 one-to-one. 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 one-to-one function, the definition
of the inverse function f –1 of f
is usually presented as:
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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 one-to-one, then it's invertible. If a function y
= f(x) isn't one-to-one, then there exist
two different x1
and x2
such that f(x1) = f(x2) = y1. Clearly its
inverse relation isn't a function, because y1 would have two different
images: x1
and x2.
So if a function isn't one-to-one then it isn't invertible. Thus a function is
invertible iff it's one-to-one.
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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(x1) = f(x2), then 2x1 + 1 = 2x2 + 1, yielding x1 = x2. So f
is one-to-one, 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
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Fig. 4.1
Graphs of y = 2x + 1 and y
= (x – 1)/2 are mirror images of |
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Fig. 4.2
Graphs of y = f(x)
and y = f –1(x)
are mirror images of each other |
domain is on the y-axis
and its range on the x-axis. Move its domain to the x-axis
and its range to the y-axis, and we get
the domain on the x-axis and the range on the y-axis 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
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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)).
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Using The Leibniz Notation
Using the Leibniz notation for ( f –1)'(x) = 1/f '( f –1(x)) we obtain dy/dx = 1/f '( y) = 1/(dx/dy):
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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.
Example 5.1
Solution
EOS
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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)).
Solution
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) = x3 + 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) = y3 + 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) = xn where x > 0 is a positive real number and n
> 0 is a positive integer.
a. Show that f –1(x) = x1/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) = yn, so that f –1(x) = y = x1/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 one-to-one differentiable function satisfying f '(x) = 1/x. Let y = f –1(x). Prove that dy/dx = y.
Solution
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