Download Inverse Functions

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MTH 125
1.5 Inverse Functions
•Recall that a function can be represented by a set of
ordered pairs. For instance, the function f (x) = x + 3
from
A = {1, 2, 3, 4} to B = {4, 5, 6, 7} can be written as
•
f : {(1, 4), (2, 5), (3, 6), (4, 7)}.
•By interchanging the first and second coordinates of
each ordered pair, you can form the inverse function
of f. This function is denoted by f –1. It is a function
from B to A, and can be written as
•
f –1: {(4, 1), (5, 2), (6, 3), (7, 4)}.
1.5
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Inverse Functions
•Note that the domain of f is equal to the range
of f –1, and vice versa, as shown in Figure 1.35.
Domain of f = range of f –1
Domain of f –1 = range of f
Figure 1.35
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Inverse Functions
•That is, when you form the composition of f
with f –1 or the composition of f –1 with f, you
obtain the identity function.
•
f(f –1(x)) = x and
f –1(f(x)) = x
Inverse Functions
•Note:
•Although the notation used to denote an
inverse function resembles exponential
notation, it is a different use of –1 as a
superscript.
•That is, in general, f –1(x) ≠ 1/f(x).
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Inverse Functions
•Here are some important observations about inverse
functions.
•1. If g is the inverse function of f, then f is the inverse
function of g.
•2. The domain of f –1 is equal to the range of f, and the
range of f –1 is equal to the domain of f.
•3. A function need not have an inverse function, but if
it does, the inverse function is unique.
•You can think of f –1 as undoing what has been
done by f.
Example 1 – Verifying Inverse Functions
•Show that the functions are inverse functions
of each other.
•f(x) = 2x3 – 1 and g(x) =
•Solution:
•Because the domains and ranges of both f and
g consist of all real numbers, you can conclude
that both composite functions exist for all x.
•The composition of f with g is given by
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Example 1 – Solution
cont’d
•
=x+1–1
•
= x.
•The composition of g with f is given by
Example 1 – Solution
•
cont’d
= x.
•Because f(g(x)) = x and g(f(x)) = x,
you can conclude that f and g are
inverse functions of each other
(see Figure 1.36).
f and g are inverse functions of each other.
Figure 1.36
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Inverse Functions
•In Figure 1.36, the graphs of f and g = f –1
appear to be mirror images of each other with
respect to the line y = x.
•The graph of f –1 is a reflection of the graph of f
in the line
y = x.
•This idea is generalized as follows.
Show analytically that f and g are inverse functions
f ( x ) = 3 − 4 x,
g ( x) =
3− x
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Existence of an Inverse Function
•Not every function has an inverse, and the
Reflective Property of Inverse Functions
suggests a graphical test for those that do—the
Horizontal Line Test for an inverse function.
•This test states that a function f
has an inverse function if and
only if every horizontal line
intersects the graph of f at most
If a horizontal line intersects the graph of f
twice, then f is not one-to-one.
once (see Figure 1.38).
Figure 1.38
Use the Horizontal Line Test to determine whether the function
is one-to-one on its entire domain and therefore has an inverse
function.
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Find the inverse function of f, graph f and f −1 on
the same set of axes, describe the relationship
between the graphs, and state the domains and
−1
ranges of f and f .
f ( x) = x 3 − 1
Inverse Trigonometric Functions
The six trig functions do not pass the horizontal line
test, so they do not have inverse functions.
To find the inverse of a trig function, you must first
restrict its domain.
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The sine function is one-to-one on the interval
[ -π/2, π/2]
y = sin x, -π/2 ≤ x ≤ π/2
Domain: [-π/2 , π/2]
Range: [-1, 1]
y = arcsin x, -1≤ x ≤ 1
Domain: [-1, 1]
Range: [-π/2 , π/2]
The cosine function is one-to-one on the
interval [0,π].
The tangent function is one-to-one on the
interval (-π/2, π/2 )
Definition of Inverse Trigonometric Function
chart on page 41
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Evaluating Inverse Trig Functions
Evaluate each of the following.
arcsin 1/2
arccos (– ½)
arctan 3
arcsin (0.8)
Properties of Inverse Trig Functions
1. sin(arcsin x) = x and arcsin(siny) = y
if -1 ≤ x ≤ 1 and –π/2 ≤ y ≤ π/2
2. tan(arctanx) = x and arctan(tany) = y
if –π/2 < y < π/2
3. sec(arcsec x) = x and arcsec(sec y) = y
if |x| ≥ 1 and 0≤ y < π/2 or π/2< y≤ π
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Using Right Triangles
Evaluate without using a calculator:
cos(arcsin 5/13)
sin θ = opp/hyp
5
13
Find the adjacent side using the Pythagorean Theorem.
θ
cos θ = adj/hyp
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