Fall 2004, Triginometry 1450-02, Week 4-5
Chapter P8. P9 Composition and Inverse of Functions
DAY 9
pp.94-109
1. General Picture
A function: y  f (x) or f : A  B .
One element from the domain A goes to only one element from the range B .
A domain A is the set of all possible values (inputs) of x for which the function is defined.
A range B is the set of all possible values (outputs) of y .
( f  g )( x)  f g ( x)
Composition () ,
(function of functions)
 f (x3 )  (x3 )2  x6
f ( x)  x
Identity function
Inverse function
(f  f
(f
EXAMPLE 6. p95
( f  g )( x)  f g ( x) 
1
1

)( x)  f f
 f )( x)  f
1
1

( x)  x
f ( x)  x 2 , f
 f ( x)   x
1
( x) 
x
Compositions of Functions
f ( x)  x  2 and g ( x)  4  x 2 . Find the values. ( f  g )( 2)
Solution:
( f  g )( x)  f ( g ( x))
1. Always write this notation
( f  g )( x)  f ( g ( x))  f (4  x 2 )
2. Then substitute g (x )
( f  g )( x)  f ( g ( x))  f (4  x )  (4  x )  2
2
2
( f  g )( 2)  (4  2 2 )  2  2
3. Then put x  g (x )
4. Put
x in the new function
Write the last answer.
Solution is: 2
REMEMBER:
1. Remember: ( f  g )( x)  f ( g ( x))
2.
Then substitute g (x ) .
3.
Then put x  g (x )
EXAMPLE 8. p109
Inverse of a Function
Find the inverse of y 
Solution:
x.
y
1. Write the function. Find
x
x  y2
y  x 2  y  f 1 ( x)  x 2
Solution is:
f 1 ( x)  x 2
2. Interchange
x from this function
x and y (after finding)
1
3. Write y  f ( x)
Try to draw the graph of 2 functions
Write the last answer.
REMEMBER:
1. Find x from the original function
2. Interchange x and y
y f
1
( x)
3.
Write

Finding the inverse means to find
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x from the original function.
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Fall 2004, Triginometry 1450-02, Week 4-5
y = x
y=x^2
y=| x |
example
ELEMENT
Identity
INVERSE
example
1, f ( x )
()
0
()
 1,  f ( x)
2, f ( x)
()
1
()
x2
Square
0, 1
Square root
Exponent
a  1, log a 1  0
Logarithm
x  axis
0 origin
y  axis
f (x)
f ( f 1 ( x))  x
a
x
f ( x)  x 2
0
f
1
2
1
f ( x)
xx
log a x
1
( x)
Range B
Domain A
,
f
1
1
2
( x)  x
2. Existence of the Inverse Function
DEFINITION:
A function: y  f (x) , f : A  B , g : B  A
Then
g  f 1 is the inverse function of f if
1.
2.
( f  g )( x)  f g ( x)  x for every x in the domain B of g and
( g  f )( x)  g  f ( x)  x for every x in the domain A of f .
A function f is one-to-one if a, b in its domain,

f (a )  f (b)  implies that a  b
A function f has an inverse function


f
1
 if and only if f is one-to-one.
To check one-to-one condition needs to check f ( a )  f (b) 
See the example 6 pp107
EXAMPLE 3. p105
ab
Verifying Inverse Functions
Show that they are inverse functions. f ( x) 
Solution:
( f  g )( x)  f g ( x)
x and g ( x)  x 2 and x  0 .
Always write this form.
f g ( x)  f ( x )  x  x  x since x  0
Substitute g (x ) and check the condition 1.
g  f ( x)   g ( x )  ( x ) 2  x
Solution is: there are inverse functions
Substitute f (x ) and check the condition 2.
Write the last answer.
2
2
REMEMBER:
1. Use ( f  g )( x)  f g ( x)
2. Check 2 conditions substituting two functions.
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Fall 2004, Triginometry 1450-02, Week 4-5
Chapter P7. P8 Exponential and Logarithmic Function
DAY 10
pp.445-464
1. General Picture
DEFINITION:
f ( x)  a x is called the exponential function with base a .
 where a  0 and a  1 . Domain: (,) Range: (0,)
f ( x)  e x is called the natural exponential function with natural base e .
 The number e  2.718 is named after the Swiss mathematician Euler (1707-1783).
e  lim (1  1n ) n . This number e is called the natural base.
n
DEFINITION:
a y  x  y  log a x
f ( x)  log a x is called the logarithmic function with base a . (Read: log base a of x )
 where a  0 and a  1 and x  0 . Domain: (0,) Range: (,)
f ( x)  log e x  ln x is called the natural logarithmic function with natural base e .
 The number e  2.718 is called Euler number.
f ( x)  log 10 x  log x is called the common logarithmic function with base 10.
 Four Arithmetic operations and Exponentation (repeated multiplication)
Operation
Addition
Subtraction
Multiplication
Division
a
ax
a x
Algebraic
ax
, where x  0
x
expressions
sum
difference
Product
Quotient
Exponentiation
ax
Exponent
p
An exponent is the power p in an expression of the form a . The process of performing the operation of
raising the base a to the power p is known as exponentiation.
 Real numbers R contain natural N , integer Z , rational Q , and irrational numbers I .
Exponent Zero
Natural exponent
Negative
Rational (radical)
Irrational
exponent
exponent
exponent
exponent
m
Number
x0
x  nN
x  n  Z x   Q
x  3I
n
a0  1
Value
a n  a  a  ...  a
(factor n times)
a n 
1
an
a
m
n
2
 (n a ) m
3
 3.32
SPECIAL PROPERTIES OF EXPONENTS
1. a  1
zero exponent
0
2. a
n

3. x  a 
n
1
an
negative exponent
x
1
n
a n a
m
n
4. a  ( n a )
Radical exponent
m
Radical exponent
a a  a
5.
Product rule
m
n
mn
mn
6. (a )  a
Power of power rule
m n
7. (a  b) m  a m  b m
Power rule
8.
am
 a mn
an
quotient rule
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Fall 2004, Triginometry 1450-02, Week 4-5
y=ln(-x)
y=x^2
y=e^x
y=x
y=ln(x)
y=sqrt(x)
Exponential and Logarithmic Function
Exponential function
Inverse function
Logarithmic function
a  x  y  log a x
a  x  y  log a x
log a x  y  a y  x
y  f ( x)  a x
a  0 , a  1, y  0
Domain: x  (,)
Range: y  (0,)
a y  x  y  log a x
y  f ( x)  log a x
a  0 , a  1 and x  0
Domain: x  (0,)
Range: y  ( ,)
y
y
Domain, range switched
1
a0  1
Zero exponent
log a 1  0
2
a a
Unit exponent
log a a  1
3
a a
4
am  an  m  n
1
m
Identity, inverse property
m
log a x  log a y  x  y
One-to-one property
a y  x  y  log a x
5. Product
rule
a m  a n  a mn
6. Quotient
rule
am
 a mn
n
a
(a m ) n  a mn
7. Power
rule
log a x  y  a y  x
() Product of 2 
() Sum of exponents
() Quotient of 
() Difference of exponents
Power of power  product
power of power
8. Change
of Base rule
log a a x  x = a loga x  x
log a ( xy)  log a x  log a y
x
 log a x  log a y
y
log a x n  n log a x
log a
Power falls down
a y  x  y  log a x
log b a  y log b a  log b x
y
Proof of 3: Let a y  x  y  log a x .. After taking the logarithm from
y  log a x 
a x  y  a x with base a ,
we get x  log a a by the definition. Also, from the definition we get a  a
x
Proof of Product: Let
log b x
log a x
y
loga x
 x.
m  log a x and n  log a y . Then a m  x and a n  y .
xy  a m  a n  a m n . After taking the logarithm with base a , log a ( xy)  m  n .
Hence, log a ( xy)  log a x  log a y .
So
Proof of Power Rule: Let
with base
m  log a x . Then a m  x . So a mn  x n . After taking the logarithm
a , log a x n  log a (a mn )  m  n using 3. Hence, log a x n  n  log a x .
Proof of Change of Base Rule: Let a y  x  y  log a x . After taking the logarithm with base
log b a y  y  log b a  log b x . Hence, y  log a x 
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b,
log b x
.
log b a
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Fall 2004, Triginometry 1450-02, Week 4-5
y=e^x
y=2^x
y=(1/10)^x
y=(1/e)^x
y=(1/2)^x
y=(0.9)^x
y=(- 0.9)^x ????
y=10^x
y=e^x
y=2^x
y=1.1^x
y=1
y=lnx/ln2
y=lnx/ln10
2. Exponential function
EXAMPLE 9. p453
Compound interest
Find the balance after 4 years if the interest is compounded (a) quarterly and (b) continuously. A total of
12000$ is invested at an annual interest rate of 3%.
Solution:
r
e  lim (1  1n ) n . For n compoundings per year: A  P (1  ) nt . For continuous compounding per
n
n
r 1
r
1
1
rt
 . Then A  P(1  ) nt  P(1  ) mrt  P((1  ) r ) rt  Pe rt .
year: A  Pe . Substitute
n m
n
m
m
r nt
a) for quarterly compoundings, n  4 . So after 4 years at 3%, the balance is A  P (1  ) =
n
0.03 44
 12.000(1 
)  13523.91
4
rt
0.034
(b) For continuous compounding, the balance is A  Pe  12000e
 13529.96  (a)
3. Logarithmic function
a y  x  y  log a x
f ( x)  log a x is called the logarithmic function with base a . (Read: log base a of x )
 where a  0 and a  1 and x  0 . Domain: (0,) Range: (,)
EXAMPLE 2, 3. p459 Logarithmic Properties
See the examples. Calculate using LOG function and try to proof general properties.
Draw the graphs.
EXAMPLE 9. p463
Domain of Logarithmic Functions
Find the Domain. f ( x)  log x 1 (2  x)
Solution: By the definition of logarithm,
1 x  2
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x  1  0  x  1 and 2  x  0  x  2 . So domain is:
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