Download Function f Function

Page 1
REVIEW (1.7)
What is an inverse function? Do all functions have inverses?
An inverse function, f -1, is a kind of “undoing” function. If the initial
function, f, takes the element a to the element b, then the inverse
function takes the element b back to the element a. The domain of f =
the range of f -1.
The range of f = the domain of f -1. The graph of f and the graph of f -1
are symmetric with respect to the line y=x. (For example, if (a,b) is on
the graph of f(x), then (b,a) is on the graph of f-1(x).
The composite of one function with its inverse becomes the identity
function. That is, if you input x in the f function “machine”, you output
f(x). If you input f(x) in the f -1 function “machine”, you output x.
f-1(f(x)) = x and also f-(f -1(x)) = x
Input x
Function
f
Output
f(x)
* Note :
Input f(x)
f −1 ( x) ≠
Function
f-1
Output
x
1
f ( x)
Page 2
Only one-to-one functions have inverses. Not all functions have
inverses because all functions are not one-to-one functions.
Definition of a one-to-one function: A function is a one-to-one if no
two different elements in the domain have the same element in the
range.
The definition of a one-to-one function can be written algebraically
as follows: A function f(x) is one-to-one
if x1 is not equal to x2 (x1 and x2 any elements of the domain)
then f(x1) is not equal to f(x2).
In other words, for any two ordered pairs (x1,y1) and (x2, y2) ,
where y1 = f(x1) and y2 = f(x2),
Then if x1 ≠ x2, then y1 ≠ y2.
Similarly, if f(x1) = f(x2), then it must be that x1 = x2.
Just as we had a vertical line test to test if a graph represents a
function, there is a horizontal line test to test if a function is 1-to-1.
Horizontal Line Test Theorem
If every horizontal line intersects the graph of a function f in at
most one point, then f is 1-to-1.
Below is the graph of y=x2-4
For y=12, there are two possible
x’s. x=-4, and x=4.
14
(-4,12)
12
(4,12)
10
8
6
4
2
0
-10
-8
-6
-4
-2
However, for each x there is
only one possible y, so y=x24 is a function.
0
2
4
6
8
10
Does not pass
Horizontal Line
Test
Therefore, this function is
not 1-to-1.
What would the inverse
fuction of y = x2 -4 be?
Solve for x.
y + 4 = x2
-2
-4
-6
A function is 1-to-1 over a certain interval
only if it is constantly decreasing or constantly
increasing over that interval. y=x2-4 is 1-to-1 over the
intervals (-∞,0) and (0, ∞)
x =± y+4
Which one do we choose?
We need to have a specific value
Page 3
Example 6 on p. 83 Finding the inverse function
Find the inverse of f(x) = 2x + 3.
Step 1: let y = f(x) and solve for y.
y = 2x + 3
y – 3 = 2x
x = y/2 – 3/2 = ½ (y-3)
Step 2: Interchange x and y and set y = f-1(x)
y = ½ (x-3)
f-1(x) = ½ (x-3) = inverse function
Step 3: Check the result by showing that
f-1(f(x)) = x and also f-(f -1(x)) = x
Plug in f(x) = 2x + 3
f -1(f(x)) = f -1(2x+3) = ½ ((2x+3) – 3) = ½ (2x + 3 – 3) = ½ (2x) = x
f-(f -1(x)) = f(½ (x-3)) = 2(½ (x-3)) + 3 = 1(x-3) + 3 = x – 3 + 3 = x
y=2x + 3
y=1/2(x-3)
y=x
10
9
8
7
6
5
4
3
2
-5
-4
-3
-2
-1
1
0
-1 0
-2
-3
-4
-5
-6
-7
-8
)=
x
f(
+
x
2
3
(1,5)
y=
1
x
f-1 (x)
2
= ½ (x-3
3
4
)
(5,1)
5
Page 4
3.1 The Inverse Sine, Cosine, and Tangent Functions
Let’s look at f(x) = sin x
The domain is all real numbers (which will represent angles).
The range is the set of real numbers where -1 ≤ sin x ≤ 1.
However, in order for the sine function to have an inverse function, it
has to be 1-to-1.
y = sin x
1.500
1.000
(π/2, 1)
(5π/2, 1)
0.500
0.000
(0,0)
Does not
Pass
Horizontal
Line Test
(π, 0)
(2π, 0)
-0.500
{-π/2 ≤x ≤ π/2} passes
-1.000
(-π/2, -1)
-1.500
Horizontal Line Test(3π/2, -1)
⎛ π π⎞
If we restrict the domain of y = sin x to the interval ⎜ − , ⎟
⎝ 2 2⎠
then it will have an inverse function. The inverse sine function is
denoted f -1 (x) = sin-1 x. The input of the inverse function is a real
number between -1 and 1, and the output of inverse sine is a real
number that is an angle in radians between –π/2 and π/2.
The domain is the set of real numbers {x| -1 ≤ x ≤ 1}.
The range is the set of real numbers such that ⎧⎨ − π2 ≤ sin x ≤ π2 ⎫⎬
⎭
⎩
By the property of inverse functions, for any x in the domain of sin x,
sin-1(sin (x)) = x and sin(sin-1(x)) = x
Notice that the input of the sine function is an angle, and the output of
the inverse sine function is an angle (in radians).
The inverse sine function also called the arcsin function.
NOTE: sin-1 x ≠ 1 / sin x
−1
Page 5
Example 3 on p.202
Find the exact value of sin-1 ( -½ )
What this is asking is, for what angle, θ, (where -π/2 ≤ θ ≤
π/2) does sin θ = ½ ?
Let’s look at the unit circle, x2 + y2 = 1. Remember each point on
the unit circle is (cos θ, sin θ). So find a point on the unit circle
with y-coordinate = -½
sin θ = - ½
when θ= -π/6
Therefore sin -1 (- ½ ) = -π/6
⎛ 2⎞
⎟
⎟
⎝ 2 ⎠
You try Exercise #7 on p.208 What is sin -1 ⎜⎜
?
Page 6
The Inverse Cosine Function
In order for the cosine function to be one-to-one we must restrict its
domain to {x| 0 ≤ x ≤ π}.
y = cos x
1.500
{0 ≤x ≤ π} passes
0, 1.000
-π/6, 0.866 π/6, 0.866
2π, 1.000
11π/6, 0.866 13π/6, 0.866
1.000
-π/3,
0.500
0.500
π/3, 0.500
-π/2, 0.000
0.000
-2π/3, -0.500
Horizontal Line Test
-0.500
-5π/6, -0.866
-1.000
5π/3, 0.500
π/2, 0.000
3π/2, 0.000
2π/3, -0.500
4π/3, -0.500
5π/6, -0.866 7π/6, -0.866
π, -1.000
7π/3, 0.500
5π/2, 0.000
8π/3, -0.500
17π/6, -0.866
3π, -1.000
-1.500
The inverse cosine function is denoted f -1 (x) = cos-1 x. The output
of the inverse cosine function is a real number that is an angle in
radians.
The domain (input) is the set of real numbers {x| -1 ≤ x ≤ 1}.
The range (output) is the set of real numbers such that {0 ≤ cos−1 x ≤ π }
By the property of inverse functions, for any x in the domain of cos x,
cos-1(cos (x)) = x and cos(cos-1(x)) = x
The inverse cosine function is also called the arccos function.
Example 6
Find the exact value of
⎛ 2⎞
⎟
cos ⎜⎜
⎟
2
⎝
⎠
−1
Look on the unit circle of an angle between 0 and π
that gives a cosine value of 2
2
Page 7
The Inverse Tangent Function
In order for the tangent function to be one-to-one we must restrict
π ⎫
⎧ π
its domain to ⎨⎩ − 2 < x < 2 ⎬⎭ . Notice that x cannot equal ± π
2
because tan x is undefined at those x-values.
2
{-π/2 <x < π/2} passes
Horizontal Line Test
1
(5π/4, 1)
(9π/4, 1)
(π/4, 1)
0
π/2
-π/2
(-π/4, -1)
-1
π
(3π/4, -1)
2π
3π/2
5π/2
(7π/4, -1)
-2
Notice that the range of the tangent function is all real numbers. The
graph extends to -∞ when close to –π/2 and to +∞ when close to π/2.
Therefore, for tan -1 x
The domain is all real numbers.
π
π
The range is the set of real numbers such that ⎧⎨ − < tan −1 x < ⎫⎬
⎩
2
2⎭
By the property of inverse functions, for any x in the domain of tan x,
tan-1(tan (x)) = x and tan(tan-1(x)) = x
Example 8 on p.207
Find the exact value of tan − 1 − 3
(
Hint, remember tan x =
)
sin x
cos x
Now you do Exercise #32 on p.208
Page 8
HOMEWORK
p. 208
Concepts and Vocab: #4,9
Exercises #3,5,11,13,15,17,25,26,29,35
Extra Credit (+2 pts)
#45 (all work must be shown)