Transcript Inverse Functions

Inverse Functions
Inverse Relations
The inverse of a relation is the set of ordered pairs
obtained by switching the input with the
output of each ordered pair in the original
relation. (The domain of the original is the range
of the inverse; and vice versa)
Ex: f  x   5x  2 and f 1  x   x5 2 are inverses because
their input and output are switched. For
instance:
f  4   22
f 1  22  4
22
4
f  x   5x  2
22
f ' x 
x2
5
4
Tables and Graphs of Inverses
Switch
x and y
Orginal
(0,25)
(20,25)
(2,16)
(18,16)
(6,4)
(14,4)
(10,0)
X
0
2
6
10
14
18
20
Y
25
16
4
0
4
16
25
X
25
16
4
0
4
16
25
Y
0
2
6
10
14
18
20
Inverse
Switch
x and y
(16,18)
(4,14)
(0,10)
(4,6)
(16,2)
Although transformed,
the graphs are identical
Line of Symmetry: y = x
Inverse and Compositions
In order for two functions to be inverses:
f
 g  x   x
AND
g  f  x   x
One-to-One Functions
A function f(x) is one-to-one on a domain D
if, for every value c, the equation f(x) = c has
at most one solution for every x in D. Or, for
every a and b in D:
f  a   f  b  unless a  b
Theorems:
1. A function has an inverse function if and only if it is
one-to-one.
2. If f is strictly monotonic (strictly increasing or
decreasing) on its entire domain, then it is one-to-one
and therefore has an inverse function.
The Horizontal Line Test
If a horizontal line
intersects a curve
more than once, it’s
inverse is not a
function.
Use the horizontal
line test to decide
which graphs have
an inverse that is a
function. Make
sure to circle the
functions.
The Horizontal Line Test
If a horizontal line
intersects a curve
more than once, it’s
inverse is not a
function.
Use the horizontal
line test to decide
which graphs have
an inverse that is a
function. Make
sure to circle the
functions.
Example
Without graphing, decide if the function
below has an inverse function.
f  x   2 x 3  6
If f is strictly monotonic (strictly increasing or decreasing) on its entire
domain, then it is one-to-one and therefore has an inverse function. See
if the derivative is always one sign:
f '  x   6 x
2
Since the derivative is always negative, the inverse of f
is a function.
Find the Inverse of a Function
1. Switch the x and y of the function whose
inverse you desire.
2. Solve for y to get the Inverse function
3. Make sure that the domains and ranges
of your inverse and original function
match up.
Example
Find the inverse of the following: d  x   4 x  3
Only Half Parabola
Switch x and y
x  4 y 3
Really y =
Solve for y
x 3  4 y
Full Parabola
x3
 y (too much)
4
2
 x3

 y
 4 
x=3
Restrict the
Domain!
 x 3
d  x  

 4 
2
1
when x  3
Make sure to check with a table and graph on the calculator.
Logarithms v Exponentials
Definition of Logarithm
The logarithm base a of b is the exponent you put
on a to get b:
log a b  x
a>0
if and only if
and
a b
b>0
x
i.e. Logs give you exponents!
The logarithm to the base e, denoted ln x, is called
the natural logarithm.
Logarithm and Exponential Forms
Logarithm Form
5 = log2(32)
Logs Give
you
Exponents
Input
Becomes
Output
Base
Stays the
Base
5
2 = 32
Exponential Form
Examples
Write each equation in exponential form
1.log125(25) = 2/3
1252/3 = 25
2.Log8(x) = 1/3
81/3 = x
Write each equation in logarithmic form
3
1.If 64 = 4
log4(64) = 3
2.If 1/27 = 3x
Log3(1/27) = x
Example
Complete the table if a is a positive real number and:
f  x  a
f  x  ax
x
f 1  x   log a x
Domain
All Reals
All Positive Reals
Range
All Positive Reals
All Reals
Continuous?
Yes
Yes
One-to-One?
Yes
Yes
Concavity
Always Up
Always Down
Left End Behavior
lim a x  0
lim  log a x   
lim a x  
lim  log a x   
Right End Behavior
x 
x 
x 0
x 
The Change of Base Formula
The following formula allows you to evaluate
any valid logarithm statement:
log c  a 
log b  a  
log c  b 
For a and b greater than 0 AND b≠1.
Example: Evaluate
log1.04  2 
ln  2 
ln 1.04 
 17.673
Solving Equations with the
Change of Base Formula
Solve:
Isolate the base
and power
2  3.46   1  909
x
2  3.46   908
x
 3.46 
Change the exponential
equation to an logarithm
equation
Use the Change of Base
Formula
x
 454
x  log3.46  454
log  454 
x
log  3.46 
x  4.9289
Properties of Logarithms
For a>0, b>0, m>0, m≠1, and any real number n.
Logarithm of 1: log m 1  0
Logarithm of the base: log m  m   1
   n  log  a 
Power Property: log m a
Product Property: logm
n
m
 a  b  logm  a   logm b
Quotient Property: logm
   logm  a   logm b
a
b
Example 1
Condense the expression:
1
3
log 5  x   log 5 10   log 5  7 
log 5  x
13
  log 10   log  7 

5

log 

5
log 5 10 x  log 5  7 
3
5
10 3 x
7
Example 2
Expand the expression:
ln  3xy
2

ln  3  ln  x   ln  y
2

ln  3  ln  x   2 ln  y 
Example 3
Solve the equation:
log 4  2   log 4  x   3
log 4  2 x   3
2x  4
3
x  32
AP Reminders
Do not forget the following relationships:
ln  e
e
x
x
ln  x 
x
e e  e
a
b
a
a b
e
a b

e
b
e
Inverse Trigonometry
Tangent Cosine
Each one of these trigonometric functions fail the
horizontal line test, so they are not one-to-one.
Therefore, there inverses are not functions.
Cosecant Secant Cotangent
Sine
Trigonometric Functions
Tangent Cosine
D :  2 , 2 
D :  0,  
D :  2 , 2 
In order for
their inverses
to be

D
:


functions, the
 2 , 0
domains of
the
trigonometric
functions are
D : 0, 2 
restricted so
that they
become oneto-one.
Cosecant Secant Cotangent
Sine
Trigonometric Functions with Restricted
Domains
 0, 2 
 2 ,  
D :  0,  
Trigonometric Functions with Restricted
Domains
Function
f (x) = sin x
f (x) = cos x
f (x) = tan x
f (x) = csc x
f (x) = sec x
f (x) = cot x
Domain
Range
 2 , 2 
 1,1
 1,1
 ,  
 ,  
 , 1 1,  
 , 1 1,  
0,  
 2 , 2 
 2 , 0   0, 2 
0, 2   2 ,  
0,  
Cos-1
Sec-1
Cot-1
Tan-1
Csc-1
Sin-1
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Function
f (x) = sin-1 x
-1
f (x) = cos x
-1
f (x) = tan x
f (x) = csc-1 x
-1
f (x) = sec x
-1
f (x) = cot x
Domain
Range
 1,1
 2 , 2 
 1,1
 ,  
 ,  
 , 1 1,  
 , 1 1,  
0,  
 2 , 2 
 2 , 0   0, 2 
0, 2   2 ,  
0,  
Alternate Names/Defintions for Inverse
Trigonometric Functions
Familiar
f (x) = sin-1 x
-1
f (x) = cos x
-1
f (x) = tan x
f (x) = csc-1 x
-1
f (x) = sec x
-1
f (x) = cot x
Alternate
Calculator
f (x) = arcsin x f (x) = sin-1 x
-1
f (x) = arccos x f (x) = cos x
-1
f (x) = arctan x f (x) = tan x
f (x) = arccsc x f (x) = sin-1 1/x
-1
f (x) = arcsec x f (x) = cos 1/x

-1
f (x) = arccot x f (x) = -tan x+2
Arccot is different because it is always positive but tan can be negative.
Example 1
Evaluate:
sin
1
 
1
2
This expression asks us to find the angle
whose sine is ½.
Remember the range of the inverse of sine
 


is  2 , 2  .
  1
Since sin    and  2  6  2 ,
6 2
1 
sin   
2 6
1
Example 2
Evaluate:
csc
1
 1
This expression asks us to find the angle
whose cosecant is -1 (or sine is -1).
Remember the range of the inverse of



,
0
0,

cosecant is  2   2  .
 
Since csc     1 and  2   2  0,
 2
csc
1
 1  

2
Example 3
tan  arcsin
Evaluate:
1
3

The embedded expression asks us to find the angle whose
sine is 1/3.
Draw a picture (There are infinite varieties):
3

1
a
Find the missing side length(s)
a 1  3
a 8 2 2
2
2
2
It does not even matter
what the angle is, we only
need to find:
tan   
opp
adj

1
2 2

2
4
Is the result positive or
negative?
Since arcsin 13  0,
tan   0
Example 4
tan  cos ( 16 ) 
1
Evaluate:
The embedded expression asks us to find the angle whose
cosine is -1/6.
Ignore the Draw a picture (There are infinite varieties):
negative for
now.
6
o

1
Find the missing side length(s)
o 1  6
o  35
2
2
2
It does not even matter
what the angle is, we only
need to find:
tan   
opp
adj

35
1
  35
Is the result positive or
negative?
Since cos
1
  16   2 ,
tan   0
Example 3
cos  tan x 
1
Evaluate:
The embedded expression asks us to find the angle whose
tangent is x.
Draw a possible picture (There are infinite varieties):
h
x

1
Find the missing side length(s)
x 1  h
2
h  x 1
2
2
2
It does not even matter
what the angle is, we only
need to find:
cos   
adj
hyp

1
x 2 1
Is the result positive or
negative?
Since - 2  tan 1 x  2 ,
cos   0
White Board Challenge
Evaluate without a calculator:
sec
1
 
2

4
White Board Challenge
Evaluate without a calculator:


3
1
cot  

3


2
3
White Board Challenge
Evaluate without a calculator:
arccos  2 x  3  
x  2
White Board Challenge
Evaluate without a calculator:
cot csc
1
 5
2 6
White Board Challenge
Evaluate without a calculator:
tan sin
1
 x 
x
2
1 x